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\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
Câu a đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot0+1=0\)
=>0x+1=0(vô lý)
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
\(\frac{2}{3}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{2}{3}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{2}{3}\)
\(x=-\frac{3}{8}\)
\(\frac{2}{3}x\)\(=\)\(\frac{1}{4}\)\(-\)\(\frac{1}{2}\)
\(\frac{2}{3}x\)\(=\)\(\frac{-1}{4}\)
\(x\)\(=\)\(\frac{-1}{4}\)\(:\)\(\frac{2}{3}\)
\(x\)\(=\)\(\frac{-3}{8}\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)
\(\Leftrightarrow x=1\)