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a, Ta có:
\(\left|2x+4\right|+\left|4x+8\right|\ge0\)
Để \(\left|2x+4\right|+\left|4x+8\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|2x+4\right|=0\\\left|4x+8\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\Rightarrow x=-2\)
Vậy...........
b, Ta có:
\(\left|x-5\right|+\left|x-7\right|\ge0\)
Để \(\left|x-5\right|+\left|x-7\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-5\right|=0\\\left|x-7\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=7\end{matrix}\right.\Rightarrow x\in\varnothing\)
Vậy...........
c,\(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\left\{{}\begin{matrix}x+8=2x+2\\x+8=-2x-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-x=-6\\3x=-10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy...........
Chúc bạn học tốt!!!
e) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-3\right)=0\) ( \(x^2+1>0\forall x\))
\(\Rightarrow x=3\)
đ) \(4.8^2=2^x\)
\(2^2.\left(2^3\right)^2=2^x\)
\(2^2.2^6=2^x\)
\(2^8=2^x\)
\(\Rightarrow x=8\)
d) \(\left|x+3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)
mấy câu trên dễ rồi tự làm em nhé
a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)
\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)
\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)
\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)
Vậy \(x=7\)
c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)
\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)
\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
\(+4x\ge6\Rightarrow|4x-6|=4x-6=|2x-4|\)
Xét tiếp:
\(+2x\ge4\Rightarrow|2x-4|=2x-4\Rightarrow4x-6-2x+4\Rightarrow2x-2=0\Rightarrow x=1\)
\(+2x\le4\Rightarrow|2x-4|=4-2x\Rightarrow4x-6=4-2x\Rightarrow4x-6-4+2x=0\Rightarrow6x-10=0\)
\(\Rightarrow x=\frac{10}{6}\)
Vậy: x=1 trong cả 2TH chỉ có 1 TH thỏa mãn
\(+4x\le6\Rightarrow|4x-6|=6-4x\)
Ta xét tiếp 2 TH sau:
\(+2x\ge4\Rightarrow|2x-4|=2x-4\Rightarrow6-4x-2x+4=0\Rightarrow10-6x=0\Rightarrow x=\frac{10}{6}\)
Bạn tự xét típ
a) 3x - 2 = 0 => 3x = 2 => x = 2/3
b) 2x - 1 = 0 => 2x = 1 => x = 1/2
c) 5 ( 4+2x) = 8+5x
<=> 20 + 10x = 8 + 5x
<=> 10x - 5x = 8 - 20
<=> 5x = -12
x = -12/5
d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)
\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)
\(\frac{31}{20}x=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)
e) 3 + 2x = 4 - 8x
<=> 2x + 8x = 4 - 3
10 x = 1
x = 1/10
f \(5+\frac{1}{2}\left(x+5\right)=3\)
\(\frac{1}{2}\left(x+5\right)=3-5=-2\)
\(x+5=-2:\frac{1}{2}=-4\)
\(x=-4-5=1\)
Vậy ......
a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Ta có : \(\left|2x+4\right|+\left|4x+8\right|=0\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|+2.\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|\left(1+2\right)=0\)
=> |2x + 4| = 0
=> 2x + 4 = 0
=> 2x = -4
=> x = -2
1. Đề đúng phải là thế này: \(\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow2x+4=4x+8=0\)
\(\Rightarrow x=-\frac{4}{2}=-\frac{8}{4}\)
\(\Rightarrow x=-2\)
2. Sửa lại đề : \(\left|x-5\right|-\left|x-7\right|=0\)
\(\Rightarrow\left|x-5\right|=\left|x-7\right|\)
\(\Rightarrow\orbr{\begin{cases}x-5=x-7\\x-5=-\left(x-7\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-5=-7\\x-5=-x+7\end{cases}}\)
( Loại trường hợp 1)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
3. \(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}x+8=2x+2\\x+8=-\left(2x+2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+8=-2x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\3x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)