0,5x =3/4x +3/4 -25%
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`M+N= 0,5x^4 -4x^3 +2x-2,5 + 2x^3 +x^2+1,5`
`= 0,5x^4 +(-4x^3+ 2x^3 ) +x^2+2x +(-2,5 +1,5)`
`= 0,5x^4 -2x^3 +x^2+2x -1`
\(M+N=0,5x^4-4x^3+2x-2,5+2x^3+x^2+1,5\)
\(=0,5x^4-4x^3+2x^3+x^2+2x-2,5+1,5\)
\(=0,5x^4-2x^3+x^2+2x-1\)
1)\(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)
2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-2x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(4-2x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x-3=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: x∈{2;3}
3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)
4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)
Ta có
0,5x=\(\frac{3}{4}x+\frac{3}{4}-\frac{1}{4}\)
0,5x=\(\frac{3}{4}x+\frac{1}{2}\)
0,5x=0,75x+1/2
0,5x= 0,5x+0,25x+1/2
0=0,25x+1/2(bớt cả hai vế cho 0,5x)
0-1/2=0,25x
-1/2=0,25x
-1/2:0,25=x
-2=x
Vậy x bằng 2
0,5x=3/4x +3/4 -25%
1/2x=3/4x+3/4-1/4
1/2x=3/4x+1/2
(1/2-3/4)x=1/2
-1/4x=1/2
x=1/2:-1/4
x=-2
\(0,5x=\frac{3}{4}x+\frac{3}{4}-25\%\)
\(\Leftrightarrow\frac{1}{2}x=\frac{3}{4}x+\frac{3}{4}-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{3}{4}x+\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow\frac{-1}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}\div\frac{-1}{4}\)
\(\Leftrightarrow x=-2\)