a) 2x(x - 5) - 2x² = 2x² - 10x - 2x² = -10x = 20 => x = 20 : (-10) = -2

b) 5x(2x - 7) + 2x(8 - 5x) = 10x² - 35x + 16x - 10x² = -19x = 5 => x = \(\frac{5}{-19}=\frac{-5}{19}\)

c) 4x(7x - 5) - 7x(4x - 2) = 28x² - 20x - 28x² + 14x = -6x = -12 => x = -12 : (-6) = 2

MK mới học lớp 7 thôi nhưng mk làm vài câu nha

a) 2x(x-5)-2x²=20

    2x²-100x-2x²=20

    100x=20

     x=20:100

      x=\(\frac{1}{5}\)

a) Ta có : 2x ( x - 5 ) - 2x² = 20

=> 2x ( x - 5 ) - 2x.x = 20

=> 2 x ( x - 5 - x ) = 20

=> 2x . ( -5 ) = 20

=> -10 x = 20

=> x = -2

Chọn mik mik làm tiếp

a: \(P=2x^2-7x+6\)

\(=2x^2-4x-3x+6\)

\(=2x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(x-2\right)\left(2x-3\right)\)

b: \(P=2x^2-7x+3\)

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

\(=2x\left(x-3\right)-\left(x-3\right)\)

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

c: \(P=2x^2+9x-5\)

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

\(=2x\left(x+5\right)-\left(x+5\right)\)

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

⇔(2 x ) ²+2.2 x .1+1 ²=0 ... c ,(3 x −4) ²−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x ²−24 x +16−126 x ²−252 x +168 x +336+147 x +294=16.

https://olm.vn/hoi-dap/detail/192758180810.html

\(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)

<=> \(-12x^2+4x+12x^2-21x=8\)

<=> \(-17x=8\)

<=> \(x=-\frac{8}{17}\)

\(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30-2x^2+5=0\)

<=> \(19x+29=0\)

<=> \(19x=-29\)

<=> \(x=-\frac{29}{19}\)

Ý cuối mình k biết -22x2 là -22.2 hay -22x² nữa :)

a)2x(x-5)-2x² = 20

\(VT=-10x\)

\(\Leftrightarrow-10x=20\)

\(\Leftrightarrow x=-2\)

b)5x(2x-7)+2x(8-5x)=5

\(VT=-19x\)

\(\Leftrightarrow-19x=5\)

\(\Leftrightarrow x=-\frac{5}{19}\)

c)4x(7x-5)-7x(4x-2)= -12

\(VT=-6x\)

\(\Leftrightarrow-6x=-12\)

\(\Leftrightarrow x=2\)

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

Thực hiện các phép đổi tương đương , ta đưa ( 1 ) về dạng :

\(\frac{x+4}{2x^2-5x+2}-\frac{x+4}{2x^2-7x+3}=0\)

\(\Leftrightarrow\left(x+4\right)\left(\frac{1}{2x^2-5x+2}-\frac{1}{2x^2-7x+3}\right)=0\)

\(\Leftrightarrow\frac{\left(x+4\right)\left(1-2x\right)}{\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)}=0\)

\(\Leftrightarrow\left(x+4\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-4\\x=\frac{1}{2}\end{array}\right.\)

Thữ vào mẫu thức : Với \(x=\frac{1}{2}\) thì \(2x^2-5x+2=0\)

Với \(x=-4\) thì \(\left(2x^2-5x+2\right)\left(2x^2-7x+3\right)\ne0\)

Vậy phương trình ( 1 ) là cho nghiệm duy nhất là \(x=-4\)

a: \(2^{x^2-2x+1}=1\)

=>\(2^{\left(x-1\right)^2}=2^0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

b: \(7^{x^2+7x}=5764801\)

=>\(7^{x^2+7x}=7^8\)

=>\(x^2+7x=8\)

=>\(x^2+7x-8=0\)

=>(x+8)(x-1)=0

=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

c: \(6^{x^2+12x}=6^{7x}\)

=>\(x^2+12x=7x\)

=>\(x^2+5x=0\)

=>x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)

=>\(3^{-x+1}=3^{2x-5}\)

=>-x+1=2x-5

=>-x-2x=-5-1

=>-3x=-6

=>x=2

e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)

=>\(5^{-3x-5}=5^{2x+1}\)

=>-3x-5=2x+1

=>-5x=6

=>\(x=-\dfrac{6}{5}\)