tìm x nguyên thỏa mãn: \(\frac{2x+2}{3x-6}=\frac{2x-6}{3x-15}\)
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\(\frac{2x}{5}+\frac{3-2x}{3}\ge\frac{3x+2}{2}\)
\(\Rightarrow\frac{12x}{30}+\frac{10\left(3-2x\right)}{30}-\frac{15\left(3x+2\right)}{30}\ge0\)
\(\Rightarrow12x+30-20x-45x-30\ge0\)
\(\Rightarrow-53x\ge0\)\(\Leftrightarrow x\le0\)\(\left(1\right)\)
\(\frac{x}{2}+\frac{3-2x}{5}\ge\frac{3x-5}{6}\)
\(\Rightarrow\frac{15x}{30}+\frac{6\left(3-2x\right)}{30}-\frac{5\left(3x-5\right)}{30}\ge0\)
\(\Rightarrow15x+18-12x-15x+25\ge0\)
\(\Rightarrow-12x\ge-43\)\(\Rightarrow12x\le43\Leftrightarrow x\le\frac{43}{12}\)\(\left(2\right)\)
Từ ( 1 ) và ( 2 ) ta có tập nghiệm chung của cả hai phương trình là \(x\le0\)
Ta có :
\(\frac{3x-2}{5}\ge\frac{x}{2}+0,8\)
\(\Leftrightarrow x\ge12\)
và \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)
\(\Leftrightarrow x< 13\) \(x\in Z\)
\(\Rightarrow x=12\)
\(\frac{3x+2}{2x^2-x-15}=\frac{3x-8}{2x^2+4x-20}\)
=>(3x+2)(2x2+4x-20)=(3x-8)(2x2-x-15)
=>6x3+16x2-52x-40=6x3-19x2-37x+120
=>16x2-52x-40=-19x2-37x+120
=>35x2=15x+160
=>7x2=3x+32
=>7x2-3x-32=0
\(\Rightarrow\left(\sqrt{7}x\right)^2-2\sqrt{7}x.\frac{3\sqrt{7}}{14}+\frac{9}{28}-\frac{905}{28}=0\)
\(\Rightarrow\left(\sqrt{7}x-\frac{3\sqrt{7}}{14}\right)^2=\frac{905}{28}\)
từ đó tìm x
\(\frac{2x+2}{3x-6}=\frac{2x-6}{3x-15}\)
\(\Rightarrow\left(2x+2\right)\left(3x-15\right)=\left(2x-6\right)\left(3x-6\right)\)
\(\Rightarrow6x^2-30x+6x-30=6x^2-12x-18x+36\)
\(\Rightarrow6x^2-30x+6x-6x^2+12x+18x=36+30\)
\(\Rightarrow6x=66\)
\(\Rightarrow x=11\)
k mk nha
\(a,\left(-2+x\right).\left(3x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Leftrightarrow}x=2}\)
Vậy ................
\(b,\left(3x+9\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+9=0\\2x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-3\end{cases}\Leftrightarrow x=-3}\)
Vậy ......................
a)\(\(\left(-2+x\right)\left(3x-6\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}-2+x=0\\3x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}\Rightarrow}x=2}\)\)
b) cmtt
_Minh ngụy_
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
Ta có \(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)
\(\Leftrightarrow\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\Leftrightarrow x+15=0\)vì \(\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)\ne0\)
\(\Leftrightarrow x=-15\)
Vậy \(x=-15\)
Ta có:\(\frac{2x+2}{3x-6}=\frac{2x-6}{3x-15}\)
\(\Rightarrow\frac{2\left(x+1\right)}{3\left(x-2\right)}=\frac{2\left(x-3\right)}{3\left(x-5\right)}\)
\(\Rightarrow\frac{2}{3}\cdot\frac{x+1}{x-2}=\frac{2}{3}\cdot\frac{x-3}{x-5}\)
\(\Rightarrow\frac{x+1}{x-2}=\frac{x-3}{x-5}\)
\(\Rightarrow\frac{x+1}{x-2}-1=\frac{x-3}{x-5}-1\)
\(\Rightarrow\frac{x+1-x+2}{x-2}=\frac{x-3-x+5}{x-5}\)
\(\Rightarrow\frac{3}{x-2}=\frac{2}{x-5}\)
\(\Rightarrow3\left(x-5\right)=2\left(x-2\right)\)
\(\Rightarrow3x-15=2x-4\)
\(\Rightarrow3x-2x=-4+15\)
\(\Rightarrow x=11\)
x=11