\((4x+y)^2-(4x-y)^2\\=[(4x+y)-(4x-y)][(4x+y)+(4x-y)]\\=(4x+y-4x+y)(4x+y+4x-y)\\=2y\cdot8x\\=16xy\)

2x - 6 - 4x = 0

-2x = -6

x = 3

Vậy x = 3

\(2(x-3)-4x=0\\\Leftrightarrow 2x-6-4x=0\\\Leftrightarrow -2x-6=0\\\Leftrightarrow -2x=6\\\Leftrightarrow x=6:(-2)\\\Leftrightarrow x=-3\)

a) \(3x+5=1-2x\)

\(\Rightarrow3x+2x=1-5\)

\(\Rightarrow5x=-4\)

\(\Rightarrow x=-\dfrac{4}{5}\)

b) \(4x\left(x-1-2x\right)2x+3=20\)

\(\Rightarrow8x^2\left(-x-1\right)+3=20\)

Bạn xem lại đề!

c) \(3x\left(2x-5\right)-4\left(5-2x\right)=0\)

\(\Rightarrow3x\left(2x-5\right)-4\left[-\left(2x-5\right)\right]=0\)

\(\Rightarrow3x\left(2x-5\right)+4\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(3x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\3x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

d) \(x^2-4x+4=25\)

\(\Rightarrow x^2-2\cdot x\cdot2+2^2=25\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(\text{#}Toru\)

Bạn nên ghi đầy đủ yêu cầu và điều kiện của đề, viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu rõ đề của bạn hơn.

\(\dfrac{x+3}{x^2-1}\cdot\dfrac{-1}{x^2+x}\)

\(=\dfrac{-\left(x+3\right)}{\left(x-1\right)\left(x+1\right)\cdot x\left(x+1\right)}\)

\(=\dfrac{-\left(x+3\right)}{\left(x-1\right)\cdot x\cdot\left(x+1\right)^2}\)

\(3+10x-x^2\\=-(x^2-10x)+3\\=-[x^2-2\cdot x\cdot5+5^2]+25+3\\=-(x-5)^2+28\)

Ta thấy: \(\left(x-5\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-5\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-5\right)^2+28\le28\forall x\)

Dấu \("="\) xảy ra khi: \(x-5=0\Leftrightarrow x=5\)

Vậy GTNN của biểu thức là 28 khi \(x=5\).

A. ( x -5 ) ( 7x + 1 ) - 7x ( x + 3)

= 7x² + x - 35x - 5 - 7x² - 21x

= (7x²-7x²) + (x - 35x - 21x) -5

= -56x - 5

B = (x² - 2x.2 + 2²) - x² + 1²

B = (x² - x²) - 4x + (2 + 1)

B= -4x +3

A. (x - 5)(7x + 1) - 7x(x + 3)

= 7x² + x - 35x - 5 - 7x² - 21x

= (7x² - 7x²) + (x - 35x - 21x) - 5

= -55x - 5

B. (x - 2)² - (x - 1)(x + 1)

= x² - 4x + 4 - x² + 1

= (x² - x²) - 4x + (4 + 1)

= -4x + 5

Bài 1:

\(a,2x^2y\left(2x^2y^2-xy^2\right)\\ =2x^2x^2y^2y-2x^2x.y^2.y=2x^4y^3-2x^3y^3\\ b,\left(x-1\right)\left(2x+3\right)\\ =x.2x+x.3-1.2x-1.3=2x^2+3x-2x-3\\ =2x^2+x-3\\ c,\left(20x^3y^4+10x^2y^3-5xy\right):5xy\\ =20x^3y^4:5xy+10x^2y^3:5xy-5xy:5xy\\ =\left(20:5\right).\left(x^3:x\right).\left(y^4:y\right)+\left(10:5\right).\left(x^2:x\right).\left(y^3:y\right)-\left(5:5\right).\left(x:x\right).\left(y:y\right)\\ =4x^2y^3+2xy^2-1\\ d,\left(y-3x\right)^2-\left(y^2-6xy\right)\\ =\left[y^2-2.y.3x+\left(3x\right)^2\right]-\left(y^2-6xy\right)\\ =y^2-6xy+9x^2-y^2+6xy =9x^2\)

Bài 2:

\(a,4xy+4xz=4x\left(y+z\right)\\ b,x^2-y^2+9-6x\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\left(x-3+y\right)\)

Bài 3:

\(a,\dfrac{3xy}{y+z}+\dfrac{3xz}{y+z}\\=\dfrac{3xy+3xz}{y+z}\\ =\dfrac{3x\left(y+z\right)}{\left(y+z\right)}=3x\left(Với:y\ne-z\right)\\ b,\dfrac{x}{x+2}-\dfrac{x}{x-2}\\ =\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x^2-2x}{\left(x+2\right)\left(x-2\right)}=0\)

a. \(x^2\) - 9y²

= (\(x\))² - (3y)²

= (\(x\) - 3y)(\(x\) + 3y)

x² - 9y² = x² - (3y)²= (x - 3y)(x + 3y)

Đề thiếu rồi. Bạn xem lại.

Bài `1`

`a, (x+2)^2 =x^2 +4x+4`

`b, (x-5)^2=x^2 -10x+25`

`c,(2x-y)^2=(2x)^2- 2*2x*y+y^2=4x^2-4xy+y^2`

`d,(3x^2-2y)^2= (3x^2)^2 - 2*3x^2*2y +(2y)^2=9x^4-12x^2y +4y^2`

`e,,(2x-y)(4x^2+2xy+y^2)= (2x)^3-y^3=8x^3-y^3`

`f,(x+3)(x^2-3x+9)=x^3+3^3=x^3+27`