Question

Bài 72 (trang 40 SGK Toán 9 Tập 1)

Phân tích thành nhân tử (với các số  $x, y, a, b$ không âm và $a \geq b$)

a) $x y-y \sqrt{x}+\sqrt{x}-1$ ;                                  b) $\sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}$ ;

c) $\sqrt{a+b}+\sqrt{a^{2}-b^{2}}$ ;                                    d) $12-\sqrt{x}-x$

Answer 1

\(a,\left(\sqrt{8}-3.\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\sqrt{8}.\sqrt{2}-3\sqrt{2}.\sqrt{2}+\sqrt{10}.\sqrt{2}-\sqrt{5}\)

\(=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}\)

\(=\sqrt{4^2}-6+\sqrt{2^2.5}-\sqrt{5}\)

\(=2-6+2\sqrt{5}-\sqrt{5}\)

\(=-2+\sqrt{5}\)

\(b,\)

\(0,2\sqrt{\left(-10^2\right).3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)

\(=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)

\(=2\sqrt{5}\)

Answer 2

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Answer 3

a) x y − y √ x + √ x − 1 = y ⋅ √ x ⋅ √ x − y √ x + √ x − 1 = y √ x ( √ x − 1 ) + ( √ x − 1 ) = ( √ x − 1 ) ( y √ x + 1 ) . b) √ a x − √ b y + √ b x − √ a y = ( √ a x + √ b x ) − ( √ a y + √ b y ) = ( √ a ⋅ √ x + √ b ⋅ √ x ) − ( √ a ⋅ √ y + √ b ⋅ √ y ) = √ x ( √ a + √ b ) − √ y ( √ a + √ b ) = ( √ a + √ b ) ( √ x − √ y ) . c) √ a + b + √ a 2 − b 2 = √ a + b + √ ( a + b ) ( a − b ) = √ a + b + √ a + b ⋅ √ a − b = √ a + b ( 1 + √ a − b ) . d) 12 − √ x − x = 12 − 4 √ x + 3 √ x − x = 4 ( 3 − √ x ) + √ x ( 3 − √ x ) = ( 3 − √ x ) ( 4 + √ x

Answer 4

a) x y-y \sqrt{x}+\sqrt{x}-1xy−yx​+x​−1

=y \cdot \sqrt{x} \cdot \sqrt{x}-y \sqrt{x}+\sqrt{x}-1=y⋅x​⋅x​−yx​+x​−1

=y \sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)=yx​(x​−1)+(x​−1)

=(\sqrt{x}-1)(y \sqrt{x}+1)=(x​−1)(yx​+1).

b) \sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}ax​−by​+bx​−ay​

=(\sqrt{a x}+\sqrt{b x})-(\sqrt{a y}+\sqrt{b y})=(ax​+bx​)−(ay​+by​)

=(\sqrt{a} \cdot \sqrt{x}+\sqrt{b} \cdot \sqrt{x})-(\sqrt{a} \cdot \sqrt{y}+\sqrt{b} \cdot \sqrt{y})=(a​⋅x​+b​⋅x​)−(a​⋅y​+b​⋅y​)

=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})=x​(a​+b​)−y​(a​+b​)

=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})=(a​+b​)(x​−y​).

c) \sqrt{a+b}+\sqrt{a^{2}-b^{2}}a+b​+a2−b2​

=\sqrt{a+b}+\sqrt{(a+b)(a-b)}=a+b​+(a+b)(a−b)​

=\sqrt{a+b}+\sqrt{a+b} \cdot \sqrt{a-b}=a+b​+a+b​⋅a−b​

=\sqrt{a+b}(1+\sqrt{a-b})=a+b​(1+a−b​).

d) 12-\sqrt{x}-x12−x​−x

=12-4 \sqrt{x}+3 \sqrt{x}-x=12−4x​+3x​−x

=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})=4(3−x​)+x​(3−x​)

=(3-\sqrt{x})(4+\sqrt{x})=(3−x​)(4+x​).

Answer 5

a) \(xy-y\sqrt{x}+\sqrt{x}-1=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(y\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

c) \(\sqrt{a+b}+\sqrt{a^2-b^2}=\sqrt{a+b}\left(1+\sqrt{a-b}\right)\)

d) \(12-\sqrt{x}-x=12-4\sqrt{x}+3\sqrt{x}-x=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)

Answer 6

Answer 7

Answer 8

a) x y-y \sqrt{x}+\sqrt{x}-1xy−yx​+x​−1

=y \cdot \sqrt{x} \cdot \sqrt{x}-y \sqrt{x}+\sqrt{x}-1=y⋅x​⋅x​−yx​+x​−1

=y \sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)=yx​(x​−1)+(x​−1)

=(\sqrt{x}-1)(y \sqrt{x}+1)=(x​−1)(yx​+1).

b) \sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}ax​−by​+bx​−ay​

=(\sqrt{a x}+\sqrt{b x})-(\sqrt{a y}+\sqrt{b y})=(ax​+bx​)−(ay​+by​)

=(\sqrt{a} \cdot \sqrt{x}+\sqrt{b} \cdot \sqrt{x})-(\sqrt{a} \cdot \sqrt{y}+\sqrt{b} \cdot \sqrt{y})=(a​⋅x​+b​⋅x​)−(a​⋅y​+b​⋅y​)

=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})=x​(a​+b​)−y​(a​+b​)

=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})=(a​+b​)(x​−y​).

c) \sqrt{a+b}+\sqrt{a^{2}-b^{2}}a+b​+a2−b2​

=\sqrt{a+b}+\sqrt{(a+b)(a-b)}=a+b​+(a+b)(a−b)​

=\sqrt{a+b}+\sqrt{a+b} \cdot \sqrt{a-b}=a+b​+a+b​⋅a−b​

=\sqrt{a+b}(1+\sqrt{a-b})=a+b​(1+a−b​).

d) 12-\sqrt{x}-x12−x​−x

=12-4 \sqrt{x}+3 \sqrt{x}-x=12−4x​+3x​−x

=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})=4(3−x​)+x​(3−x​)

=(3-\sqrt{x})(4+\sqrt{x})=(3−x​)(4+x​).

Answer 9

a) x y-y \sqrt{x}+\sqrt{x}-1xy−yx​+x​−1

=y \cdot \sqrt{x} \cdot \sqrt{x}-y \sqrt{x}+\sqrt{x}-1=y⋅x​⋅x​−yx​+x​−1

=y \sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)=yx​(x​−1)+(x​−1)

=(\sqrt{x}-1)(y \sqrt{x}+1)=(x​−1)(yx​+1).

b) \sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}ax​−by​+bx​−ay​

=(\sqrt{a x}+\sqrt{b x})-(\sqrt{a y}+\sqrt{b y})=(ax​+bx​)−(ay​+by​)

=(\sqrt{a} \cdot \sqrt{x}+\sqrt{b} \cdot \sqrt{x})-(\sqrt{a} \cdot \sqrt{y}+\sqrt{b} \cdot \sqrt{y})=(a​⋅x​+b​⋅x​)−(a​⋅y​+b​⋅y​)

=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})=x​(a​+b​)−y​(a​+b​)

=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})=(a​+b​)(x​−y​).

c) \sqrt{a+b}+\sqrt{a^{2}-b^{2}}a+b​+a2−b2​

=\sqrt{a+b}+\sqrt{(a+b)(a-b)}=a+b​+(a+b)(a−b)​

=\sqrt{a+b}+\sqrt{a+b} \cdot \sqrt{a-b}=a+b​+a+b​⋅a−b​

=\sqrt{a+b}(1+\sqrt{a-b})=a+b​(1+a−b​).

d) 12-\sqrt{x}-x12−x​−x

=12-4 \sqrt{x}+3 \sqrt{x}-x=12−4x​+3x​−x

=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})=4(3−x​)+x​(3−x​)

=(3-\sqrt{x})(4+\sqrt{x})=(3−x​)(4+x​).

Answer 10

(\(^{^{ }}\sqrt{x}\)+ \(\dfrac{2}{3}\)+\(\sqrt{x^2-y^2}\)

Answer 11

máy em bị lỗi

Answer 12

bbb

Answer 13

 

 

Answer 14

 

Answer 15

 

Answer 16

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Answer 17

a)(\(\sqrt{x}\)-1)(y\(\sqrt{x}\)+1) với x≥1
b)(\(\sqrt{a}\)+\(\sqrt{b}\))(\(\sqrt{x}\)-\(\sqrt{y}\)) với a,b,x,y > 0
c)(\(\sqrt{a+b}\))(1+\(\sqrt{a-b}\))
d)(4+\(\sqrt{x}\))(3-\(\sqrt{x}\))

Answer 18

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Answer 19

 

Answer 20

a) xy - y√x + √x - 1

= (√x)2.y - y√x + √x - 1

= y√x(√x - 1) + √x - 1

= (√x - 1)(y√x + 1) với x ≥ 1

= √x(√a + √b) - √y(√a + √b)

= (√a + √b)(√x - √y) (với x, y, a và b đều không âm)

(với a + b, a - b đều không âm)

d) 12 - √x - x

= 16 - x - 4 - √x (tách 12 = 16 - 4 và đổi vị trí)

= [42 - (√x)2] - (4 + √x)

= (4 - √x)(4 + √x) - (4 + √x)

= (4 + √x)(4 - √x - 1)

= (4 + √x)(3 - √x)

Answer 21

a,\(\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)với x≥1

b,\(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)với x,y,a,b >0

c,\(\sqrt{a+b}\left(1+\sqrt{a-b}\right)\)với a+b,a-b>0

d,\(\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)

Answer 22

a) x y-y \sqrt{x}+\sqrt{x}-1xy−yx​+x​−1

=y \cdot \sqrt{x} \cdot \sqrt{x}-y \sqrt{x}+\sqrt{x}-1=y⋅x​⋅x​−yx​+x​−1

=y \sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)=yx​(x​−1)+(x​−1)

=(\sqrt{x}-1)(y \sqrt{x}+1)=(x​−1)(yx​+1).

b) \sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}ax​−by​+bx​−ay​

=(\sqrt{a x}+\sqrt{b x})-(\sqrt{a y}+\sqrt{b y})=(ax​+bx​)−(ay​+by​)

=(\sqrt{a} \cdot \sqrt{x}+\sqrt{b} \cdot \sqrt{x})-(\sqrt{a} \cdot \sqrt{y}+\sqrt{b} \cdot \sqrt{y})=(a​⋅x​+b​⋅x​)−(a​⋅y​+b​⋅y​)

=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})=x​(a​+b​)−y​(a​+b​)

=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})=(a​+b​)(x​−y​).

c) \sqrt{a+b}+\sqrt{a^{2}-b^{2}}a+b​+a2−b2​

=\sqrt{a+b}+\sqrt{(a+b)(a-b)}=a+b​+(a+b)(a−b)​

=\sqrt{a+b}+\sqrt{a+b} \cdot \sqrt{a-b}=a+b​+a+b​⋅a−b​

=\sqrt{a+b}(1+\sqrt{a-b})=a+b​(1+a−b​).

d) 12-\sqrt{x}-x12−x​−x

=12-4 \sqrt{x}+3 \sqrt{x}-x=12−4x​+3x​−x

=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})=4(3−x​)+x​(3−x​)

=(3-\sqrt{x})(4+\sqrt{x})=(3−x​)(4+x​).

Answer 23

a)

 $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

 

b)

 $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

 

c)

 $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

 

d)

 $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Answer 24

a) (\(\sqrt{x}-1\) ) ( y\(\sqrt{x}\)+ 1)

b) ( \(\sqrt{a}+\sqrt{b}\)) (\(\sqrt{x}-\sqrt{y}\))

c) \(\sqrt{a+b}\) (1+\(\sqrt{ }\)(a-b))

d) (3-\(\sqrt{x}\)) (4+\(\sqrt{x}\))

Answer 25

 

Answer 26

 

Answer 27

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

Answer 28

 

Answer 29

a)\(xy-y\sqrt{x}+\sqrt{x}-1\)

=\(y\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)

=\(\left(\sqrt{x}-1\right)\left(y+1\right)\)

b)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

=\(\sqrt{ax}-\sqrt{ay}-\left(\sqrt{by}-\sqrt{bx}\right)\)

=\(\sqrt{a}.\sqrt{x}-\sqrt{a}.\sqrt{y}-\left(\sqrt{b}.\sqrt{y}-\sqrt{b}.\sqrt{x}\right)\)

=\(\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\left(\sqrt{b}\left(\sqrt{y}-\sqrt{x}\right)\right)\)

=\(\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)

=\(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}\sqrt{b}\right)\)

c)\(\sqrt{a+b}+\sqrt{a^2-b^2}\)

=

 

Answer 30