$x-3)2=9$

⇒[2x−3=32x−3=−3⇒[x=3x=0⇒[2x−3=32x−3=−3⇒[x=3x=0

Vậy x = 3 hoặc x = 0

\(\left(2x-3\right)^2=9\)

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

⇒\(2x-3=+-3\)

\(TH1:2x-3=3\text{⇒}x=3\)

\(TH2:2x-3=-3\text{⇒}x=0\)

ta có: \(\left(2x-3\right)^2\)=\(3^2\)

2x-3=3

2x=3+3

2x=6

x=6:2

x=3

vậy x==3

b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)

\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)

\(\Leftrightarrow2x^2-4x-2x^2-6x-3x+3-29=0\)

\(\Leftrightarrow-13x-26=0\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=26:-13\)

\(\Leftrightarrow x=-2\)

Vậy ...

\(x\left(2x-4\right)-2x\left(x+3\right)-3\left(x-1\right)-29=0\)

\(2x^2-4x-2x^2-6x-3x+3-29=0\)

\(2x^2-4x-2x^2-6x-3x=0+29-3\)

\(\left(2x^2-2x^2\right)+\left(-4x-6x-3x\right)=26\)

\(0+\left(-4-6-3\right)x=26\)

\(\Rightarrow-13x=26\rightarrow x=-2\)

(2x-123)-(x+27)=0

2x-123-x-27=0

x-150=0

x=150

Ukm

It's very hard

l can't do it 

Sorry!

a) \(x^4-x^3-7x^2+x+6=0\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt

b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)

\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)

Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

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

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

\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)

Từ đó tính đc x

d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+5=a\), khi đó pt có dạng:

\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

\(|2x-7|=12\Leftrightarrow\orbr{\begin{cases}2x-7=12\\2x-7=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=19\\2x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{19}{2}\\x=-\frac{5}{2}\end{cases}}}\)

\(|4x+3|=|3x-1|\Leftrightarrow\orbr{\begin{cases}4x+3=3x-1\\4x+3=1-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\7x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{2}{7}\end{cases}}}\)

\(|3x+5|=2x+9\left(ĐKXĐ:2x+9\ge0\Leftrightarrow x\ge-\frac{9}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=2x+9\\3x+5=-2x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\5x=-14\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\left(tm\right)\\x=-\frac{14}{5}\end{cases}}}\left(tm\right)\)

Tự KL cho mỗi phần

a)\(\text{ 4x - 15 = -75 - x}\)

\(4x-15+75+x=0\)

\(5x+60=0\)

\(5x=-60\)

\(x=-14\)

Vậy....

Thêm dấu suy ra trc mỗi dòng nha

Học tốt

b)\(|2x-7|+2=13\)

\(|2x-7|=11\)

\(\Leftrightarrow\hept{\begin{cases}2x-7=11\\2x-7=-11\end{cases}\Leftrightarrow\hept{\begin{cases}2x=18\\2x=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\x=2\end{cases}}}\)

vậy x=9 hoặc x=2

a) 4x - 15 = -75 - x

4x + x = -75 + 15

5x = -60

x= -60:5

x=-12

b) |2x - 7| + 2 = 13

| 2x - 7 | = 11

th1: 2x-7= 11

2x = 18

x = 9

th2: 2x - 7 = -11

2x = -4

x = -2

c)(2x - 1)^2 = 9

(2x-1)^2= 3^2

2x-1 = 3

2x = 4

x = 2

d)2x (x - 3) = 0

Th1: 2x=0

x= 0

Th2: x- 3 =0

x= 3

Hok tốt!!!