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|7 + 5x| = 1 - 4x
=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)
=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)
|4x2 - 2x| + 1 = 2x
=> |4x2 - 2x| = 2x - 1
=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)
Vậy ...
a)\(\sqrt{5}\left(x+2\right)=\sqrt{10}\)
\(\Leftrightarrow x+2=\sqrt{2}\)
\(\Leftrightarrow x=\sqrt{2}-2\)
a/ \( (a+1)x^2−2(a+3)x+2 =0\) (1)
Với \(a+1=0\Leftrightarrow a=-1\) thì \(\left(1\right)\Leftrightarrow-4x+2=0\Leftrightarrow x=\frac{1}{2}\)
Với \(a\ne-1\), ta có: \(\Delta'=\left(a+3\right)^2-2\left(a+1\right)=a^2+4a+7=\left(a+2\right)^2+3>0\forall x\in R\)
Suy ra ĐPCM
b/ \(x^ 2 +(a+1)x+2(a^ 2 −a+1) =0\)
có \(\Delta=\left(a+1\right)^2-4.2\left(a^2-a+1\right)=-7a^2+10a-7\)
Đề sai bạn nhé, vì phương trình có thể vô nghiệm nha bạn!
1: =>x+1=5
=>x=4
2: \(\Leftrightarrow\left|x-5\right|=2x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x+2-x+5\right)\left(2x+2+x-5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+7\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
3: \(\Leftrightarrow\sqrt{3+x}\left(\sqrt{3-x}+1\right)=0\)
=>x+3=0
=>x=-3
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) M\(=\frac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(=\frac{x}{\sqrt{x}+1}\)
b) Khi \(x=7+4\sqrt{3}\Rightarrow\frac{7+4\sqrt{3}}{\sqrt{\left(2+\sqrt{3}\right)^2}+1}=\frac{7+4\sqrt{3}}{3+\sqrt{3}}\)
c)\(M=\frac{1}{2}\Leftrightarrow\frac{x}{\sqrt{x}+1}=\frac{1}{2}\Leftrightarrow\sqrt{x}=2x-1\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x^2=4x^2-4x+1\Leftrightarrow3x^2-4x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=\frac{1}{3}\left(l\right)\\x=1\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
d. x+165+x+363=x+561+x+759
\(\Leftrightarrow2x+528=2x+1320\)
\(\Leftrightarrow2x-2x=1320-528\)
\(\Leftrightarrow0x=729\) (loại)
\(\Rightarrow\) PT vô N0