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\(\frac{2x-10}{6}=\frac{-27}{5-x}\)
\(\frac{2\left(x-5\right)}{6}=\frac{27}{x-5}\)
\(2\left(x-5\right)^2=27\times6\)
\(2x^2-20x+50-162=0\)
\(2x^2-20x-112=0\)
\(x^2-10x-56=0\)
\(x^2-14x+4x-56=0\)
\(\left(x-14\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-14=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)
KL .....
XIN TIICK
a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
Vậy ...
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6}=\frac{2x+1+3y-2-2x-3y+1}{5+7-6}=\frac{0}{6}=0\)
\(\Rightarrow2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2};\)
\(3y-2=0\Rightarrow3y=2\Rightarrow y=\frac{2}{3}\)
Vậy \(x=-\frac{1}{2};y=\frac{2}{3}\)
Áp dụng tc cua dtsbn ta có
\(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}=\frac{2x+1+3y-2}{5+7}=\frac{2x+3y-1}{12}\left(1\right)\)
\(\Rightarrow\frac{2x+3y-1}{6x}=\frac{2x+3y-1}{12}\Rightarrow6x=12\Rightarrow x=2\)
Thay vào 1 ta có:\(\frac{2.2+1}{5}=\frac{3y-2}{7}\Rightarrow1=\frac{3y-2}{7}\Rightarrow\frac{3y-2}{7}=1\)
\(\Rightarrow3y-2=7\Rightarrow3y=9\Rightarrow y=3\)
Vậy.....
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
I 2x-3 I = I x+1 I
2x-3 = x+1
x+1 - 2x+3=0
x (1-2) +1+3=0
-1x +4 =0
-1x = 0-4
-1x =-4
x = -4 : -1
x =4
Trả lời:
\(\left|2x-3\right|=\left|x+1\right|\)
\(\Rightarrow2x-3=x+1\) hoặc \(2x-3=-\left(x+1\right)\)
TH1: \(2x-3=x+1\)
\(2x-x=1+3\)
\(x=4\)
TH2: \(2x-3=-\left(x+1\right)\)
\(2x-3=-x-1\)
\(2x+x=-1+3\)
\(3x=2\)
\(x=\frac{2}{3}\)
Vậy \(x=4;x=\frac{2}{3}\)
\(\frac{13x-2}{2x+5}=-\frac{27}{5-x}\)ĐK : \(x\ne-\frac{5}{2};5\)
\(\Rightarrow\left(13x-2\right)\left(5-x\right)=-27\left(2x+5\right)\)
\(\Leftrightarrow65x-13x^2-10+2x=-54x-135\)
\(\Leftrightarrow-13x^2+121x+125=0\Leftrightarrow x=10,24...;x=-0,93...\)