a) x² - 4x - 5 = 0

=> x² - 5x + x - 5 = 0

=> x(x - 5) + (x - 5) = 0

=> (x + 1)(x - 5) = 0

=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

b) 4x² + 7x - 11 = 0

=> 4x² + 11x - 4x - 11 = 0

=> x(4x + 11) - (4x + 11) = 0

=> (x - 1)(4x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)

c) -7x² + 6x + 1 = 0

=> -7x² + 7x - x + 1 = 0

=> -7x(x - 1) - (x - 1) = 0

=> (-7x - 1)(x - 1) = 0

=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)

d) -10x² + 7x + 3 = 0

=> -10x² + 10x - 3x + 3 = 0

=> -10x(x - 1) - 3(x - 1) = 0

=> (-10x - 3)(x - 1) = 0

=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(a,x^2-4x-5=0\)

\(\Rightarrow x^2-5x+x-5=0\)

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

\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(b,4x^2+7x-11=0\)

\(\Rightarrow4x^2-4x+11x-11=0\)

\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)

\(c,-7x^2+6x+1=0\)

\(\Rightarrow-7x^2+7x-x+1=0\)

\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

\(d,-10x^2+7x+3=0\)

\(\Rightarrow-10x^2+10x-3x+3=0\)

\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)

a) 4x²  - 2x + 3 - 4x.(x - 5) = 7x - 3

--> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

--> 4x² - 2x - 4x² + 20x - 7x = -3 - 3

--> 11x = -6

--> x = \(\frac{-6}{11}\)

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

--> -3x² + 15x + 5x - 5 + 3x² = 4x

--> -3x²  + 15x + 5x + 3x² - 4x = 5 

--> 16x = 5

--> x = \(\frac{5}{16}\)

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

--> 7x² - 14x - 5x + 5 = 7x² + 3 

--> 7x²  - 14x - 5x - 7x²  = -5 + 3 

--> -19x = -2 

--> x = \(\frac{2}{19}\)

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

--> 15x - 3 - x² + 2x + x² - 13x = 7

--> 15x - x² + 2x + x² - 13x = 3 + 7

--> 4x = 10

--> x = \(\frac{5}{2}\)

e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12

--> 2x² - 3x - 2x² + 10x = 12

--> 7x = 12

--> x = \(\frac{12}{7}\)

~ Học tốt ~

a) 4x² - 2x + 3 - 4x(x - 5) = 7x - 3

=> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

=> 18x + 3 = 7x - 3

=> 18x - 7x = -3 - 3

=> 11x = -6

=>  x = -6/11

b) -3x(x - 5) + 5(x - 1) + 3x² = 4x

=> -3x² + 15x + 5x - 5 + 3x² = 4x

=> 20x - 5 = 4x

=> 20x - 4x = 5

=> 16x = 5

=> x = 5/16

\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)

\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)

\(\Leftrightarrow7x^2-7x^2-19x=3-5\)

\(\Leftrightarrow-19x=-2\)

\(\Leftrightarrow x=\frac{2}{19}\)

a) 4x² - 2x + 3 - 4x.(x - 5) = 7x - 3

<=> 18x + 3 = 7x - 3

<=> 18x = 7x - 3 - 3

<=> 18x = 7x - 6

<=> 18x - 7x = -6

<=> 11x = -6

<=> x = -6/11

=> x = -6/11

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

<=> 20x - 5 = 4x

<=> 20x = 4x + 5

<=> 20x - 4x = 5

<=> 16x = 5

<=> x = 5/16

=> x = 5/16

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

<=> 7x.(x - 2) - 5.(x - 1) = 7x² + 3

<=> 7x² - 19x + 5 = 7x² + 3

<=> 7x² - 19x = 7x² + 3 - 5

<=> 7x² - 19x = 7x² - 2 

<=> 7x² - 19x - 7x² = -2

<=> -19x = -2

<=> x = 2/19

=> x = 2/19

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

<=> 4x - 3 = 7

<=> 4x = 7 + 3

<=> 4x = 10

<=> x = 10/4

=> x = 5/2

e) 1/5x.(10x - 15) - 2x.(x - 5) = 12

<=> x(2x - 3) - 2x(x - 5) = 12

<=> 7x = 12

<=> x = 12/7

=> x = 12/7

a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)

Bài 1: 

a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)

\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)

\(\Rightarrow16x-5=x-2\)

\(\Rightarrow16x-x=5-2\)

\(\Rightarrow15x=3\)

\(\Rightarrow x=\dfrac{15}{3}=5\)

b) \(12x^2-4x\left(3x+5\right)=10x-17\)

\(\Rightarrow12x^2-12x^2-20x=10x-17\)

\(\Rightarrow-20x=10x-17\)

\(\Rightarrow-20x-10x=-17\)

\(\Rightarrow-30x=-17\)

\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)

c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)

\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)

\(\Rightarrow-8x=12\)

\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)

Bài 2: 

a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)

\(=x^2-7x+5x-35-7x^2+21x\)

\(=-6x^2+19x-35\)

b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)

\(=x^3-x^2-2x-x^2+x-5x-5\)

\(=x^3-2x^2-6x-5\)

c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)

\(=x^2-7x-5x+35-x^2-3x+4x-12\)

\(=11x+23\)

d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)

\(=x^2-2x-x+2-x^2+2x+5x+10\)

\(=4x+12\)

Bài 2: 

a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)

\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)

\(=\left(7x+5+3x-5\right)^2\)

\(=\left(10x\right)^2=100x^2\)

Thay x=-2 vào A, ta được:

\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)

b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)

\(=8x^3+y^3-8x\left(x^2-1\right)\)

\(=8x^3+y^3-8x^3+8x\)

\(=8x+y^3\)

Thay x=-2 và y=3 vào B, ta được:

\(B=-2\cdot8+3^3=-16+27=11\)

Ai help mk vs

Bài 1: 

Ta có: \(A=x\left(4-x\right)\)

\(=4x-x^2\)

\(=-\left(x^2-4x\right)\)

\(=-\left(x^2-4x+4\right)+4\)

\(=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{max}=4\) khi x=2

a) 5.(x^2-3x+1)+x.(1-5x)=x-2

\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)

\(\Leftrightarrow-14x-x=-2-5\)

\(\Leftrightarrow-15x=-7\)

\(\Leftrightarrow x=\frac{7}{15}\)

b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)

\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)

\(\Leftrightarrow-4x^2+15x-10=0\)

Đề sai???

\(c,12x^2-4x\left(3x-5\right)=10x-17\)

\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)

\(\Leftrightarrow10x=-17\)

\(\Leftrightarrow x=-\frac{17}{10}\)

\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)

\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)

\(\Leftrightarrow8x=12\)

\(\Leftrightarrow x=\frac{3}{2}\)

Bạn cứ phân tích hết ra nhé

Không bt mk ms hỏi chứ  nếu phân tích đc mk đã phân tích gòi

phần b phân tích đc = 12x²+20x-12x²+18x-38x+5 = 5

=> BT ko phụ thuộc vào giá trị của biến