\(X\cdot4+0,5\cdot X=55,35\)

\(X\cdot\left(4+0,5\right)=55,35\)

\(X\cdot4,5=55,35\)

\(X=55,35:4,5\)

\(X=12,3\)

\(A\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\\\Leftrightarrow \left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\\ \Leftrightarrow\left(x-1\right)^{x+2}\left(\left(x-1\right)^{x+2}+1\right)=0\\ \Leftrightarrow\left(x-1\right)^{x+2}=0hoac\left(x-1\right)^{x+2}+1=0\)

Giả tiếp đc x=1

a: \(=12x^2-9x-12x^2-10x+6x+5=-13x+5\)

b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x\)

\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)

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

c: \(=x^3-3x^2+3x-1+x^3+8+3\left(x^2-16\right)\)

\(=2x^3-3x^2+3x+7+3x^2-48=2x^3+3x-41\)

d: \(=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\)

Bài 1:

a) Ta có: \(A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)

\(=2x^2+6x-2\left(x^2-2x-8\right)\)

\(=2x^2+6x-2x^2+4x+16\)

\(=10x+16\)

Thay \(x=\frac{1}{2}\) vào biểu thức \(A=10x+16\), ta được:

\(A=10\cdot\frac{1}{2}+16=5+16=21\)

Vậy: 21 là giá trị của biểu thức \(A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right)\) tại \(x=\frac{1}{2}\)

b) Ta có: \(B=\left(3x+4\right)^2-\left(x+4\right)\left(x+4\right)-10x\)

\(=9x^2+24x+16-\left(x^2+8x+16\right)-10x\)

\(=9x^2+24x+16-x^2-8x-16-10x\)

\(=8x^2+6x\)

Thay \(x=\frac{1}{10}\) vào biểu thức \(B=8x^2+6x\), ta được:

\(B=8\cdot\left(\frac{1}{10}\right)^2+6\cdot\frac{1}{10}=8\cdot\frac{1}{100}+\frac{6}{10}\)

\(=\frac{8}{100}+\frac{6}{10}\)

\(=\frac{8}{100}+\frac{60}{100}=\frac{17}{25}\)

Vậy: \(\frac{17}{25}\) là giá trị của biểu thức \(B=\left(3x+4\right)^2-\left(x+4\right)\left(x+4\right)-10x\) tại \(x=\frac{1}{10}\)

c) Ta có: \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)

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

\(=x^2+2x+1-4x^2+4x-1+3x^2-12\)

\(=6x-12\)

Thay x=1 vào biểu thức C=6x-12, ta được:

\(C=6\cdot1-12=6-12=-6\)

Vậy: -6 là giá trị của biểu thức \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\) tại x=1

d) Ta có: \(D=\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2-2x\left(x-4\right)\)

\(=x^2-9+x^2-4x+4-2x^2+8x\)

\(=4x-5\)

Thay x=-1 vào biểu thức D=4x-5,ta được:

\(D=4\cdot\left(-1\right)-5=-4-5=-9\)

Vậy: -9 là giá trị của biểu thức \(D=\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2-2x\left(x-4\right)\) tại x=-1

a: =>2x-2/3-3x+3/2=1/2x

=>-3/2x=-5/6

=>x=5/6:3/2=5/6x2/3=10/18=5/9

b: =>-3x+3/4-1/3x-1/6=x

=>-13/3x=-7/12

=>x=7/12:13/3=7/12x3/13=21/156=7/52

chiều tau bày cho

Bài 3: Tìm x, biết:

a) \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow4x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10

b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)

\(\Leftrightarrow2\left(1-2x\right)=18\)

\(\Leftrightarrow2-4x=18\)

\(\Leftrightarrow4x=-16\)

\(\Leftrightarrow x=-4\)

Vậy x =-4

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

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

\(\Leftrightarrow x=\frac{5}{12}\)

Vậy \(x=\frac{5}{12}\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow25-6x=9\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\)

Vậy \(x=\frac{8}{3}\)

f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow21-5x=0\)

\(\Leftrightarrow5x=21\)

\(\Leftrightarrow x=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

bài của bạn làm sai rồi :)

a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)

b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)

e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)