Giải phương trình: (x+5) / (x+10) = 2/3
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![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(2x-1+5\left(3-x\right)>0\\ 2x-2+15-5x>0\\ -3x+13>0\\ x< \dfrac{13}{3}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2< 10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-10< 0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+4x-1-10< 0\)
\(\Leftrightarrow-20x< -25\)
\(\Leftrightarrow x>\dfrac{5}{4}\)
\(b,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)\le3\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)\le3\)
\(\Leftrightarrow x^3-25x-\left(x^3+8\right)\le3\)
\(\Leftrightarrow x^3-25x-x^3-8-3\le0\)
\(\Leftrightarrow-25x\le11\)
\(\Leftrightarrow x\ge-\dfrac{11}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
=>(4x+14+3x+9)(4x+14-3x-9)=0
=>(7x+23)(x+5)=0
=>x=-23/7 hoặc x=-5
\(a,\\ \Leftrightarrow7x^2+58x+115=0\\ \Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
\(b,\\ \Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=0\\ \LeftrightarrowĐặt.x^2+6x+5=a\\ \Leftrightarrow a=a\left(a+3\right)=10\\ \Leftrightarrow a^2+3a-10=0\\ \Leftrightarrow\left(a+5\right)\left(a-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-5\\a=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+6x+5=-5\\x^2+6x+5=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+6x+10=0\\x^2+6x+3=0\end{matrix}\right.\\ \left(Vô.n_o\Delta=36-40=-4< 0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\)