7,      4x mũ 2 - 12x + 9 - y mũ 2 =  -(y-2x+3) (y+2x-3)

8,      16x mũ 2 - 4y mũ 2 + 4y - 1 =   -(2y - 4x - 1) (2y+4x-1)

9,        25 - x mũ 2 - 12x - 36 =  -(x+1) (x+11)

10,        x mũ 2 - 9 - 5 ( x + 3 ) =  (x-8) (x+3)

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chúc bạn học tốt :))))

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Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM 

Trả lời:

7, 4x² - 12x + 9 - y²

= ( 4x² - 12x + 9 ) - y² 

= ( 2x - 3 )² - y² 

= ( 2x - 3 - y )( 2x - 3 + y )

8, 16x² - 4y² + 4y - 1

= 16x² - ( 4y² - 4y + 1 )

= ( 4x )² - ( 2y - 1 )²

= ( 4x - 2y + 1 )( 4x + 2y - 1 )

9, 25 - x² - 12x - 36

= 25 - ( x² + 12x + 36 )

= 5² - ( x + 6 )²

= ( 5 - x - 6 )( 5 + x + 6 )

= ( - 1 - x )( 11 + x )

10, x² - 9 - 5 ( x + 3 )

= ( x - 3 )( x + 3 ) - 5 ( x + 3 )

= ( x + 3 )( x - 3 - 5 )

= ( x + 3 )( x - 8 )

6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)

7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)

8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)

9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)

10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)

mk mới làm r nha bn

1. \(4x^2-49=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

===========

2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

===========

3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\)

hay x=6

a)  2x-3=-x+6

2x - 3 + x - 6 =0

3x -9 = 0

3x = 9

x = 9 : 3

x= 3

c/ \(\left(-12x-4^3\right).8^3=4.8^4\)

\(\left(-12x-64\right).512=16384\)

\(-12x-64=\dfrac{16384}{512}=32\)

\(-12x=32+64=96\)

\(x=\dfrac{96}{\left(-12\right)}=-8\)

a,   2x-3=-x+6

\(\Leftrightarrow\)2x-3+x-6=0

\(\Leftrightarrow\)3x-9=0

\(\Leftrightarrow\)3x=9

\(\Leftrightarrow\)x=3

a) Ta có: \(2x-3=-x+6\)

\(\Leftrightarrow2x+x=6+3\)

\(\Leftrightarrow3x=9\)

hay x=3

Vậy: x=3

b) Ta có: \(\left|2x-7\right|+2=13\)

\(\Leftrightarrow\left|2x-7\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=11\\2x-7=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{9;-2\right\}\)

c) Ta có: \(\left(-12x-4^3\right)\cdot8^3=4\cdot8^4\)

\(\Leftrightarrow\left(-12x-64\right)=\dfrac{4\cdot8^4}{8^3}=32\)

\(\Leftrightarrow-12x=96\)

hay x=-8

Vậy: x=-8

a) Ta có: \(x^2-x+2\)

\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\)

Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall x\)

hay \(x^2-x+2>0\forall x\)

Vậy: Đa thức \(x^2-x+2\) vô nghiệm(đpcm)

b) Ta có: \(4x^2-12x+10\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2+1\)

\(=\left(2x-3\right)^2+1\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x-3\right)^2+1\ge1>0\forall x\)

hay \(4x^2-12x+10>0\forall x\)

Vậy: Đa thức \(4x^2-12x+10\) vô nghiệm(đpcm)

a: \(\Leftrightarrow A=25x^2y-13xy^2+y^3-11x^2y+2y^3=14x^2y-13xy^2+3y^3\)

b: \(A=-12x^4+15x^2y-2xy^2-7\)

a. Theo đề bài ta có :

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\) và 5x-y+3z=124

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=31\Rightarrow x=31.3=93\\\dfrac{y}{5}=31\Rightarrow y=31.5=155\\\dfrac{z}{-2}=31\Rightarrow z=\left(-2\right).31=-62\end{matrix}\right.\)

Vậy.........

\(a,x:y:z=5:3:\left(-2\right)\)và \(5x-y+3z=124\)

Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{\left(-2\right)}\Rightarrow\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{\left(-6\right)}\)

\(=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)

\(\Rightarrow x=31.3=93\)

\(y=31.5=155\)

\(z=31.\left(-2\right)=\left(-62\right)\)

Vậy........

\(b,x:y=3:7\)và \(x^2-2y^2=\left(-356\right)\)

Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x^2}{9}=\dfrac{2y^2}{98}\)

\(\dfrac{x^2-2y^2}{9-98}=\dfrac{\left(-356\right)}{\left(-89\right)}=4\)

\(\Rightarrow x=4.3=12\)

\(y=4.7=28\)

Vậy.......