\(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-8x+16=4\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy...

\(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow x^2-13x+22=0\)

\(\Leftrightarrow\left(x+\frac{13}{2}\right)^2=\frac{81}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-21}{2}\end{cases}}\)

Vậy...

b/ \(x^2-4=8\left(x-2\right)\)

<=> \(x^2-4=8x-16\)

<=> \(x^2-4-8x+16=0\)

<=> \(x^2-8x+12=0\)

<=> \(x^2-8x+16-4=0\)

<=> \(\left(x-4\right)^2-4=0\)

<=> \(\left(x-4\right)^2=4\)

<=> \(\orbr{\begin{cases}x-4=2\\x-4=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=2\end{cases}}\)

c/ \(x^2-4x+4=9\left(x-2\right)\)

<=> \(\left(x-2\right)^2-9\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(x-2-9\right)=0\)

<=> \(\left(x-2\right)\left(x-11\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x=11\end{cases}}\)

d/ \(4x^2-12x+9=\left(5-x\right)^2\)

<=> \(\left(2x-3\right)^2-\left(5-x\right)^2=0\)

<=> \(\left(2x-3-5+x\right)\left(2x-3+5-x\right)=0\)

<=> \(\left(3x-8\right)\left(x+2\right)=0\)

<=> \(\orbr{\begin{cases}3x-8=0\\x+2=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=8\\x=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)

a)4×(x-5)-(x-1)×(4x-3)=5

=>4x-20-4x²+7x-3-5=0

=>-4x²+11x-28=0

=>-4(x²-\(\frac{11x}{4}\)+7)=0

=>\(-4\left(x-\frac{11}{8}\right)^2-\frac{327}{16}< 0\)

=>vô nghiệm

b) (3x-4)(x-2)=3x(x-9)-3

=>3x²-10x+8=3x²-27x-3

=>17x=-11

=>x=-11/17

c)(x-5)×(x-4)-(x+1)×(x-2)=7

=>x²-9x+20-x²+x+2=7

=>22-8x=7

=>-8x=-15

=>x=8/15

d)(2x-1)×(x-2)-(x+3)×(2x-7)=3

=>2x²-5x+2-2x²+x+21=3

=>23-4x=3

=>-4x=-20

=>x=5

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)

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

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

a) 3(x - 1)² - 3x(x - 5) = 2

=> 3(x² - 2x + 1) - 3x² + 15x = 2

=> 3x² - 6x + 3 - 3x² + 15x = 2

=> 9x = 2 - 3

=> 9x = -1

=> x = -1/9

b) 4x² - 12x = -9

=> 4x² - 12x + 9 = 0

=> (2x - 3)² = 0

=> 2x - 3 = 0

=> 2x = 3

=>  x = 3/2

c) (2x - 3)² = (x  + 5)²

=> (2x - 3)² - (x + 5)² = 0

=> (2x - 3 - x - 5)(2x - 3 + x + 5) = 0

=> (x - 8)(3x + 2) = 0

=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

d) \(\left(x^4-2x^3+4x^2-8x\right):\left(x^2+4\right)-2x=-4\)

=> \(\left[x^3\left(x-2\right)+4x\left(x-2\right)\right]:\left(x^2+4\right)-2x=-4\)

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

=> \(x^2-2x-2x+4=0\)

=> \(\left(x-2\right)^2=0\)

=> x - 2 = 0

=> x = 2

e) khđ

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)

\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)

\(\Leftrightarrow5x-3x+4x=35-15-48+48\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)

\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)

\(\Leftrightarrow6x-28x-18x=30+32-5-9\)

\(\Leftrightarrow-40x=48\)

\(\Leftrightarrow x=-1.2\)

C,D nưã bạn

 c, 7 . ( - x - 7 ) - 5 . ( - x - 3 ) = 12 . ( 3 - x)

<=> - 7x - 49  + 5x  + 15 = 36 - 12x

<=>  - 2x + 12x = 36 - 15 + 49

<=> 10 x = 70

<=> x = 7

Vẫy  x = 7

d, 5 . ( - 3x - 7 ) - 4 . ( - 2x - 11 ) = 7 . ( 4x + 10) + 9

<=> - 15x + 8x + 44 = 28x + 70 + 9

<=> - 7x + 44 = 28x + 79

<=> - 7x - 28x = 79 - 44

<=> - 35x = 35

<=> x = - 1

Vậy x = - 1

Hc tốt

a. x=1

b. x=1

a. x = 1

b. x = 1

a,x=\(\frac{4}{3}\)

b, x= 1

c, x= viết đề rõ hơn

d, \(\frac{13}{6}\)

rút gọn biểu thức

a) \(4x^2-\left(x+3\right).\left(x-5\right)+x\)

\(=4x^2-\left(x^2-5x+3x-15\right)+x\)

\(=4x^2-x^2+5x-3x+15+x\)

\(=3x^2+3x+15.\)

b) \(x.\left(x-5\right)-3x.\left(x+1\right)\)

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

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

\(=-2x^2-8x.\)

d) \(\left(x+3\right).\left(x-1\right)-\left(x-7\right).\left(x-6\right)\)

\(=x^2-x+3x-3-\left(x^2-6x-7x+42\right)\)

\(=x^2-x+3x-3-x^2+6x+7x-42\)

\(=15x-45.\)

Chúc bạn học tốt!

Lời giải:

a. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}=4+3.\sqrt{\frac{1}{9}}.\sqrt{x-5}$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

b. Sửa đoạn 4x-45 thành 4x-20.

ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{4}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$

$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=\frac{12}{5}$

$\Leftrightarrow x-5=\frac{144}{25}=5,76$

$\Leftrightarrow x=10,76$ (tm)

a: \(\Leftrightarrow\left|x-1\right|=3-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3-x+1\right)\left(2x+3+x-1\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(3x+2\right)=0\\x< =\dfrac{3}{2}\end{matrix}\right.\)

=>x=-2/3

b: Trường hợp 1: x<-3

Pt sẽ là:

\(-x-1-x-3=10-4x\)

=>-2x-4=10-4x

=>2x=14

hay x=7(loại)

Trường hợp 2: -3<=x<-1

Pt sẽ là \(x+3-x-1=10-4x\)

=>10-4x=2

=>4x=8

hay x=2(loại)

Trường hợp 3: x>=-1

Pt sẽ là x+1+x+3=10-4x

=>2x+4=10-4x

=>6x=6

hay x=1(nhận)