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a) \(||2x-3|-4x|=5\)
TH1: \(|2x-3|-4x=5\)
\(\Leftrightarrow|2x-3|=5+4x\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5+4x\\2x-3=-5-4x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-4x=5+3\\2x+4x=-5+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=8\\6x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{-1}{3}\end{cases}}\)
TH2: \(|2x-3|-4x=-5\)
\(\Leftrightarrow|2x-3|=-5-4x\)<0 ( loại )
Vậy \(x\in\left\{-4;\frac{-1}{3}\right\}\)
b) \(\left||3x+1|+3\right|=2\)
Mà \(\left|3x+1\right|\ge0\)nên \(\left|3x+1\right|+3\ge3\)
Vậy biểu thức trong dấu GTTĐ luôn dương
\(\Rightarrow\left|3x+1\right|+3=2\)
\(\Rightarrow\left|3x+1\right|=-1\)(vô lí)
Vậy pt vô nghiệm
a) \(\left|2x-1\right|-4=5\)
\(\Leftrightarrow\left|2x-1\right|=5+4\)
\(\Leftrightarrow\left|2x-1\right|=9\)
\(\Leftrightarrow2x-1=\pm9\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
c) \(\left|3x-2\right|=4-2x\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=4-2x\\-\left(3x-2\right)=4-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-2\end{cases}}\)
d) \(\left|1-3x\right|=1+2x\)
\(\Leftrightarrow\orbr{\begin{cases}1-3x=1+2x\\-\left(1-3x\right)=1+2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
a) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
=> \(\frac{3}{2}x-\frac{2}{5}-\frac{1}{3}x+\frac{1}{4}=0\)
=> \(\left(\frac{3}{2}-\frac{1}{3}\right)x+\left(-\frac{2}{5}+\frac{1}{4}\right)=0\)
=> \(\frac{7}{6}x-\frac{3}{20}=0\)
=> \(\frac{7}{6}x=\frac{3}{20}\)
=> \(x=\frac{3}{20}:\frac{7}{6}=\frac{3}{20}\cdot\frac{6}{7}=\frac{9}{70}\)
b) \(2x-\frac{2}{3}=7x+\frac{2}{3}-1\)
=> \(2x-\frac{2}{3}=7x-\frac{1}{3}\)
=> \(2x-\frac{2}{3}-7x+\frac{1}{3}=0\)
=> (2x - 7x) + (-2/3 + 1/3) = 0
=> -5x - 1/3 = 0
=> -5x = 1/3
=> x = -1/15
a)\(A=x^2-1\)
\(Nx:\)\(x^2\ge0\)
\(\Rightarrow A_{Min}=0-1=-1\Leftrightarrow x=0\)
b) \(B=x^2-2x+3\)
\(=x\left(x-2\right)+3\)
\(Nx:x\left(x-2\right)\ge0\)
\(\Rightarrow B_{Min}=3\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow x=0\)
c) \(C=\left|2x+1\right|-5\)
\(Nx:\left|2x+1\right|\ge0\Rightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
\(\Rightarrow C_{Min}=-5\Leftrightarrow x=\frac{-1}{2}\)
d) \(D=3x^2+6x-7\)
\(=3\left(x^2+2x\right)-7\)
\(Nx:Min_{x^2+2x}=-1\Leftrightarrow x=-1\)
\(D_{Min}=-8\Leftrightarrow x=-1\)
a) B = | 2x - 3 | - 7
| 2x - 3 | ≥ 0 ∀ x => | 2x - 3 | - 7 ≥ -7
Đẳng thức xảy ra <=> 2x - 3 = 0 => x = 3/2
=> MinB = -7 <=> x = 3/2
C = | x - 1 | + | x - 3 |
= | x - 1 | + | -( x - 3 ) |
= | x - 1 | + | 3 - x | ≥ | x - 1 + 3 - x | = | 2 | = 2
Đẳng thức xảy ra khi ab ≥ 0
=> ( x - 1 )( 3 - x ) ≥ 0
=> 1 ≤ x ≤ 3
=> MinC = 2 <=> 1 ≤ x ≤ 3
b) M = 5 - | x - 1 |
- | x - 1 | ≤ 0 ∀ x => 5 - | x - 1 | ≤ 5
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MaxM = 5 <=> x = 1
N = 7 - | 2x - 1 |
- | 2x - 1 | ≤ 0 ∀ x => 7 - | 2x - 1 | ≤ 7
Đẳng thức xảy ra <=> 2x - 1 = 0 => x = 1/2
=> MaxN = 7 <=> x = 1/2
a) \(\left(1-2x\right)^3=-8\)
\(\left(1-2x\right)^3=\left(-2\right)^3\)
\(1-2x=-2\)
\(2x=1-\left(-2\right)\)
\(2x=3\)
\(x=3:2\)
\(x=1,5\)
b) \(\left(2x-1\right)^3=-27\)
\(\left(2x-1\right)^3=\left(-3\right)^3\)
\(2x-1=-3\)
\(2x=-3+1\)
\(2x=-2\)
\(x=-2:2\)
\(x=-1\)
@Nghệ Mạt
#cua
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5