3(x-2)-4(2x+1)-5(2x+3)=50
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\(\left(2x-5\right)\left(x-3\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(x-3+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(3x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\3x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{8}{3}\end{cases}}\)
\(\frac{3x-5}{4}+\frac{2x-3}{6}=\frac{x}{3}-1\)
\(\Leftrightarrow\frac{18x-30+8x-12}{24}=\frac{x-3}{3}\)
\(\Leftrightarrow\frac{26x-42}{24}=\frac{x-3}{3}\)
\(\Leftrightarrow78x-126=24x-72\)
Chuyển vế các kiểu
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\(3\left(x-2\right)-4\left(2x+1\right)-5\left(2x+3\right)=50\)
<=>\(3x-6-8x-4-10x-15=50\)
<=>\(-15x-25=50\)
<=>\(-15x=75\)
<=>\(x=-5\)
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a: \(\Leftrightarrow2x^2-6x+x-3=5x+4-2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2-5x-3=5x+4-2x^2+8\)
\(\Leftrightarrow4x^2-10x-9=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot4\cdot\left(-9\right)=100+144=244>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{10-\sqrt{244}}{8}=\dfrac{5-\sqrt{61}}{4}\\x_2=\dfrac{5+\sqrt{61}}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow3x-4x^2-4x-1=4x^2-\left(2x^2+2x-x-1\right)\)
\(\Leftrightarrow-4x^2-x-1-4x^2+2x^2+x-1=0\)
\(\Leftrightarrow-6x^2-2=0\)
hay \(x\in\varnothing\)
c: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{x^2+2x-3}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=x^2+2x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+2x-7\)
=>5x+2=2x-7
=>3x=-9
hay x=-3(loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-2\right)-4\left(2x+1\right)-5\left(2x+3\right)=50\)
\(\Rightarrow3x-6-8x-4-10x-15=50\)
\(\Rightarrow3x-8x-10x=50+6+4+15\)
\(\Rightarrow3x-8x-10x=75\)
\(\Rightarrow x\left(3-8-10\right)=75\)
\(\Rightarrow x.\left(-15\right)=75\)
\(\Rightarrow x=75:\left(-15\right)\)
\(\Rightarrow x=-5\)
\(\text{Vậy }x=-5\)
Ta có:3(x-2) - 4(2x+1) - 5(2x+3)=50
=>3x-6-8x-4-10x-15=50
=>-15x-25=50
=>-15x=75
=>x=-5
Vậy x=-5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:x\ne-1;x\ne\dfrac{1}{2}\)
Ta có : \(\dfrac{6x+5}{3x+3}=\dfrac{5-4x}{1-2x}\)
\(\Leftrightarrow\left(6x+5\right)\left(1-2x\right)=\left(5-4x\right)\left(3x+3\right)\)
\(\Leftrightarrow6x-12x^2+5-10x=15x+15-12x^2-12x\)
\(\Leftrightarrow5-4x-12x^2-15-3x+12x^2=0\)
\(\Leftrightarrow-10-7x=0\)
\(\Rightarrow x=\dfrac{-10}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
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