Đăng từng câu đio

,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn

(x+1/5) -4 = -2

x= -6-1/5

x= -31/5

Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)

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

Tìm x,biết :

\(a,\left(x+\frac{1}{5}\right)-4=-2\) 

\(\left(x+\frac{1}{5}\right)=2\)  

\(x+\frac{1}{5}=2\) 

\(x=\frac{9}{5}\) 

b,\(\left(2x+3\right)\left(x-7\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\) 

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\) 

\(c,\frac{31}{9}x-\frac{5}{2}=\frac{8}{3}\) 

\(\frac{31}{9}x=\frac{8}{3}+\frac{5}{2}\) 

\(\frac{31}{9}x=\frac{31}{6}\) 

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

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

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

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

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

a) \(5^{x+2}-5^{x-1}=3100\) \(\Leftrightarrow5^x.5^2-5^x:5=3100\)

\(\Leftrightarrow5^x.25-5^x.\frac{1}{5}=3100\)\(\Leftrightarrow5^x.\left(25-\frac{1}{5}\right)=3100\)

\(\Leftrightarrow5^x.\frac{124}{5}=3100\)\(\Leftrightarrow5^x=125=5^3\)\(\Leftrightarrow x=3\)

Vậy \(x=3\)

b) \(\left(x-4\right)\left(2x+3\right)< 0\)

TH1: \(\hept{\begin{cases}x-4>0\\2x+3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\2x< -3\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< \frac{-3}{2}\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-4< 0\\2x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\2x>-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x>\frac{-3}{2}\end{cases}}\Leftrightarrow\frac{-3}{2}< x< 4\)

mà x là số nguyên \(\Rightarrow-1< x< 4\)

Vậy \(-1< x< 4\)

a, \(\left|2x+1\right|=5\Rightarrow2x+1\in\left\{5;-5\right\}\)

+) Nếu :\(2x+1=5\Rightarrow2x=4\Rightarrow x=4\div2=2\)

+) Nếu : \(2x+1=-5\Rightarrow2x=-6\Rightarrow x=-6\div2=-3\)

Vậy \(x\in\left\{2;-3\right\}\)

b, \(\left|x-4\right|=\left|2-x\right|\)

\(\Rightarrow\left[\begin{matrix}x-4=2-x\\x-4=-\left(2-x\right)\end{matrix}\right.\)

+) Nếu : x - 4 = 2 - x

\(\Rightarrow x+x=2+4\Rightarrow2x=6\Rightarrow x=3\)

+) Nếu : x - 4 = - ( 2 - x )

\(\Rightarrow x-4=-2+x\Rightarrow x-x=-2+4\Rightarrow0=2\) ( loại )

Vậy x = 3 thỏa mãn đề bài

c, \(\left|x-5\right|=2-x\Rightarrow\left|x-5\right|+x=2\)

+) Nếu : \(x< 5\Rightarrow x-5< 5-5\Rightarrow x-5< 0\Rightarrow\left|x-5\right|=-x+5\)

Thay vào đề , ta có :

\(-x+5+x=2\Rightarrow-x+x+5=2\Rightarrow5=2\) ( loại )

+) Nếu : \(x\ge5\Rightarrow x-5\ge5-5\Rightarrow x-5\ge0\Rightarrow\left|x-5\right|=x-5\)

Thay vào đề , ta có :

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

\(\Rightarrow x-x-5=2\Rightarrow-5=2\) ( loại )

Vậy \(x\in\varnothing\)

\(2\left(|x-1|+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(TH1:x\ge1\Rightarrow|x-1|=x-1\)

\(\Rightarrow2\left(x-1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(\Rightarrow2\left(2x-\frac{9}{5}\right)=2x-\frac{2}{5}\Rightarrow4x-\frac{18}{5}=2x-\frac{2}{5}\)

\(\Rightarrow4x-2x=\frac{18}{5}-\frac{2}{5}\Rightarrow2x=\frac{16}{5}\Rightarrow x=\frac{16}{5}:2=\frac{16}{10}=\frac{8}{5}\)

\(TH2:x< 1\Rightarrow|x-1|=-x+1\)

\(\Rightarrow2\left(-x+1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(\Rightarrow2\left(1-\frac{4}{5}\right)=2x-\frac{2}{5}\Rightarrow2\cdot\frac{1}{5}=2x-\frac{2}{5}\)

\(\Rightarrow2x-\frac{2}{5}=\frac{2}{5}\Rightarrow2x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\Rightarrow x=\frac{4}{5}:2=\frac{4}{10}=\frac{2}{5}\)