a) Cho a/b =b/c=c/d . CMR (a+b+c /b+c+d)2 = a/d
b) Tìm x biết |17x -5 | - |17x + 5 | =0
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10 tháng 5 2018
\(\text{a) }\left|\left|x+5\right|-4\right|=3\)
- Xét \(x\ge-5\Leftrightarrow\left|x+1\right|=3\):
+) Với \(x\ge-1\Leftrightarrow x+1=3\)
\(\Leftrightarrow x=2\left(T/m\right)\)
+) Với \(-5\le x< -1\Leftrightarrow-x-1=3\)
\(\Leftrightarrow x=-4\left(T/m\right)\)
- Xét \(x< -5\Leftrightarrow\left|x-9\right|=3\)
+) Với \(-5< x< 9\Leftrightarrow9-x=3\)
\(\Leftrightarrow x=6\left(T/m\right)\)
+) Với \(x\ge9\left(loại\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{2;-4;6\right\}\)
\(\text{b) }\left|17x-5\right|-\left|17x+5\right|=0\\ \Leftrightarrow\left|17x-5\right|=\left|17x+5\right|\\ \Leftrightarrow\left[{}\begin{matrix}17x-5=\left(17x+5\right)\\17x-5=-\left(17x+5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}17x-5=17x+5\\17x-5=-17x-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}17x-17x=5+5\\17x+17x=-5+5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=10\left(loại\right)\\34x=0\end{matrix}\right.\Leftrightarrow x=0\)
Vậy phương trình có nghiệm \(x=0\)
\(\text{c) }\left|3x+4\right|=2\left|2x-9\right|\\ \Leftrightarrow\left[{}\begin{matrix}3x+4=2\left(2x-9\right)\\3x+4=-2\left(2x-9\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+4=4x-18\\3x+4=-4x+18\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x-4x=-18-4\\3x+4x=18-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-22\\7x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=22\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{2;22\right\}\)
6 tháng 8 2017
a) \(2x^2+x-6=0\Leftrightarrow2x^2+4x-3x-6=0\)
\(\Leftrightarrow2x\left(x+2\right)-3\left(x+2\right)=0\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
vậy \(x=\dfrac{3}{2};x=-2\)
b) \(-5x^2+17x-6=0\Leftrightarrow-5x^2+15x+2x-6=0\)
\(\Leftrightarrow-5x\left(x-3\right)+2\left(x-3\right)\Leftrightarrow\left(-5x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5x+2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\)
vậy \(x=\dfrac{2}{5};x=3\)
c) \(3x^2+22x-16=0\Leftrightarrow3x^2+24x-2x-16=0\)
\(\Leftrightarrow3x\left(x+8\right)-2\left(x+8\right)=0\Leftrightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\x=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
vậy \(x=\dfrac{2}{3};x=-8\)
d) \(2x^3+3x^2-8x+3=0\Leftrightarrow2x^3-3x^2+x+6x^2-9x+3=0\)
\(\Leftrightarrow x\left(2x^2-3x+1\right)+3\left(2x^2-3x+1\right)=0\Leftrightarrow\left(x+3\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x^2-2x-x+1\right)=0\Leftrightarrow\left(x+3\right)\left(2x\left(x-1\right)-\left(x-1\right)\right)=0\)
\(\left(x+3\right)\left(2x-1\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\2x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2x=1\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\) vậy \(x=-3;x=\dfrac{1}{2};x=1\)
a) Sửa đề CMR : \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)
=> \(\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\left(\frac{a}{b}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\left(\frac{a+b+c}{b+c+d}\right)^3\)
=> \(\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(\text{vì }\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\right)\)
=> \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(\text{đpcm}\right)\)
b) |17x - 5| - |17x + 5| = 0
=> |17x - 5| = |17x + 5|
=> \(\orbr{\begin{cases}17x-5=17x+5\\17x-5=-17x-5\end{cases}}\Rightarrow\orbr{\begin{cases}0x=10\\34x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\in\varnothing\\x=0\end{cases}}\Rightarrow x=0\)
Vậy x = 0 là giá trị cần tìm