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[Chapter 21]

P3 exercise:

1, solution: Let the length and width of this rectangle be 2x cm and 3x cm respectively.

From the meaning of the title, it is 2x 3x = 18. Since x 2 = 3, then x=√3 or -√3 (omitted),

So the length and width of this rectangle are 2√3 cm and 3√3 cm respectively.

2. Solution: From the figure, AC=2, AB=3, AC⊥AB.

According to Pythagorean theorem, BC 2 = AB 2+AC 2 = 2 2+3 2 =13,

∴BC=√ 13.

3. Solution: (1) According to the meaning of the question, a- 1≥0, that is, a≥ 1.

(2) According to the meaning of the question, 2a+3≥0, then a≥-3/2.

P5 exercise:

1, (1) Original formula =3. (2) The original formula =9×2= 18.

2.( 1) Original formula =0.3. (2) The original formula = 1/7. (3) The original formula =-π. (4) The original formula = √1102 =1/.

P8 exercise:

1, (1) Original formula =√ 10. (2) The original formula =6. (3) Original formula =2√y. (4) Original formula =2.

2.( 1) Original formula =77. (2) The original formula = 15. (3) original formula =2√y. (4) original formula =4|bc|√ac.

3. Solution: The area of the rectangle is s = √ 10× 2 √ 2 = 4 √ 5 (cm 2).

P 1 1 exercise:

1, solution: (1) Original formula =3. (2) The original formula =2√3. (3) The original formula =√3/3. (4) The original formula =2|a|.

2.( 1) Original formula =4√2. (2) The original formula =2√ 10. (3) The original formula = 6/2. (4) The original formula =2√3/3.

3. solution: let BC=x cm.

Angle c = 90, angle a = 30,

∴AB=2BC=2x cm.

∫ab 2 = BC 2+AC 2，

∴(2x)^2=4+x^2.

The solution is x 1=2√3/3, x2=-2√3/3 (abbreviated),

That is, AB=4√3/3 cm.

P 16 exercise:

1, solution: (1) is incorrect, √8-√3=2√2-√3. (2) Incorrect, √4+√9=2+3=5. (3) correct.

2. Solution: (1) Original formula =-4√7. (2) The original formula =4√5-2√5+√5=3√5.

(3) The original formula = 3 √ 2+(7 √ 2-3 √ 3) = 10 √ 2-3. (4) The original formula = (2 √ 6+2/2)-(2/4-6) = 3 √ 6+√.

3. Let the radius of the big circle be R cm and the radius of the small circle be r cm.

From π r 2 = 25. 12, π r 2 = 12.56,

R=√25. 12/3. 14=2√2，

r=√ 12.56/3. 14=2，

Then d= R-r=2√2-2≈0.83(cm).

P 17 exercise:

1, solution: (1) Original formula =√6+√ 10. (2) The original formula =4+2√2. (3) The original formula = 1 1+5√5. (4) original.

2. Solution: (1) Original formula =9. (2) The original formula =4. (3) The original formula =7+4√3. (4) The original formula =22-4√ 10.

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[Chapter 22]

P27 exercise:

1, solution: (1) The original equation can be transformed into 5x 2-4x- 1 = 0. The quadratic coefficient is 5; The coefficient of the first term is -4,; The constant term is-1.

(2) The original equation can be changed to 4x 2-8 1 = 0. The quadratic coefficient is 4; The coefficient of the first term is 0; The constant term is -8 1.

(3) The original equation can be changed to 4x 2+8x-25 = 0. The quadratic coefficient is 4; The coefficient of the first term is 8; The constant term is -25.

(4) The original equation can be changed to 3x 2-7x+ 1 = 0. The quadratic coefficient is 3; The coefficient of the first term is-7; The constant term is 1.

2. Solution: (1) According to the meaning of the question, it is 4x 2 = 25, which is converted into the general formula 4x 2-25 = 0.

(2) According to the meaning of the question, x(x-2)= 100 is converted into the general formula x 2-2x- 100 = 0.

(3) According to the meaning of the question, x 1 = (1-x) 2, and the general formula is x 2-3x+1= 0.

P28 exercise:

1, solution:-2,3 is the root of the equation x 2-x-6 = 0.

2. Solution: 2,0 and 1 are the roots of the equation x 2-x = 0.

P3 1 exercise:

Solution: (1) The original equation can be simplified as x 2 = 4, x 1=2, and x2=-2.

(2) The original equation can be simplified as 9x 2 = 8, x 1 = 2 √ 2/3, x2=-2√2/3.

(3) The original equation can be changed to (x+6) 2 = 9, x 1=-3, x2=-9.

(4) The original equation can be changed to (x- 1) 2 = 2, x 1= 1+√2, x2= 1-√2.

(5) The original equation can be changed to (x-2) 2 = 5, x 1=2+√5, x2=2-√5.

(6) The original equation can be changed to (3x+ 1) 2 = 4, x 1= 1/3, x2 =- 1.

P34 Exercise:

1、( 1)5^2 5 (2)6^2 6 (3)(5/2)^2 5/2 (4) 1/3.

2. Solution: (1) The original equation can be changed to (x+5) 2 = 16, then x 1=- 1, x2=-9.

(2) The original equation can be changed to (X- 1/2) 2 = 2, then x 1=( 1+2√2)/2, and x2=( 1-2√2)/2.

(3) The original equation can be changed to (x+ 1) 2 = 7/3, then x 1=(-3+√2 1)/3, x2=(-3-√2 1)/3.

(4) The original equation can be changed to (x-3/4) 2 = 2116, then x 1=(3+√2 1)/4, x2 = (3-√ 2/kloc-0.

(5) The original equation can be changed to (x+ 1) 2 =- 1 < 0, then this equation has no real root.

(6) The original equation can be changed to (x-2) 2 = 16, then x 1=-2 and x2=6.

P37 Exercise:

1 、( 1)x 1=2，x2=-3。 (2) x 1=(√3+2)/2，x2=(√3-2)/2。

(3)x 1=(3+√ 15)/3，x2=(3-√ 15)/3。 (4)x 1=3/2，x2=0。

(5)x 1=√3，x2=-√3。 (6)x 1=(-2+√ 14)/2，x2=(-2-√ 14)/2。

2. solution: for the equation x 2-75x+350 = 0, a= 1, b=-75, c=350,

Then x = 75 √ (-75) 2-4× 350/2 = 75 65/2.

That is, x 1=5 and x2=70 (excluding). Then x=5.

P40 exercise:

1, solution: (1) The equation can be changed to x(x+ 1)=0. Then x 1=- 1, x2=0.

(2) The equation can be changed to x(x-2√3)=0. Then x 1 = 2 √ 3, x2 = 0.

(3) The equation can be changed to 3 (x- 1) 2 = 0. Then x 1=x2= 1.

(4) The equation can be changed to (2x-11) (2x+11) = 0. Then x 1 = 1 1/2, x2 =- 166.

(5) The equation can be changed to 6x2-x-2 = 0. Then x 1=- 1/2, x2=2/3.

(6) The equation can be changed to (3x = 9) (-x+ 1) = 0. Then x 1= 1, x2=3.

2. solution: let the radius of the small circle be r m.

According to the meaning, π r 1 = 1/2 π (r+5) 2.

Then 2r 2 = (r+5) 2.

The solution is r 1=5+5√2, r2=5-5√2 (omitted).

The radius of the small circular site is (5+5√2) meters.

P24 exercise:

Solution: (1)x 1+x2=3, x 1 x2 =- 15.

(2)x 1+x2= 1，x 1 x2=- 1。

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I'm exhausted, that's all! ^_^

Hope to adopt ~