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a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)
vậy phương trình vô nghiệm
a) (x-1)2 = 0
<=> x-1 = 0
<=> x = 1
b) (x-2)2 - 1 = 0
<=> (x-2)2 = 1
<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) (2x-1)3 = -8
<=> (2x-1)3 = -23
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(-\dfrac{1}{2}\)
d) (x+2)2 + 1 = 0
<=> (x+2)2 = -1
<=> x+2 = -1
<=> x = -3
1)
\(2^{x-1}=16\\ 2^{x-1}=2^4\\ \Rightarrow x-1=4\\ x=4+1\\ x=5\)
5)
\(\left(x-1\right)^2=25\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6)
\(\left|2x-1\right|=5\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
5) (x-1)2 = 25
(x-1)2 = 52
x-1 = 5
x = 5+1
x = 6
6) \(\left|2x-1\right|=5 \)
\(TH1:\) \(2x-1=5\)
\(\Leftrightarrow2x=5+1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\)
\(TH2:2x-1=-5\)
\(\Leftrightarrow2x=-5+1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4:2\)
\(\Leftrightarrow x=-2\)
Vậy x = 3 hoặc x = -2.
Tick nha!
2x-3y+5z=1 hoặc =-1
TH1: \(\dfrac{x}{y}\)=\(\dfrac{3}{2}\)=>\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\)=>\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)
\(\dfrac{y}{z}\)=\(\dfrac{5}{7}\)=>\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)=>\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)
\(\Rightarrow\)\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)=>\(\dfrac{2x}{30}\)=\(\dfrac{3y}{30}\)=\(\dfrac{5z}{70}\)
Áp dụng tính chát dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-3y+5z}{30-30+70}\)=\(\dfrac{1}{70}\)
=>x=1.15:7=\(\dfrac{3}{14}\)
y=\(\dfrac{1}{7}\)
z=\(\dfrac{1}{5}\)
TH2:............=-1 tự tính nhé làm tương tựmình còn phải ôn bài
a) Xem lại đề
b) Ta có: \(2x=4y=5z\)=> \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}\) => \(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}=\frac{2x-3y-z}{1-\frac{3}{4}-\frac{1}{5}}=\frac{1}{\frac{1}{20}}=20\)
=> \(\hept{\begin{cases}\frac{x}{\frac{1}{2}}=20\\\frac{y}{\frac{1}{4}}=20\\\frac{z}{\frac{1}{5}}=20\end{cases}}\) => \(\hept{\begin{cases}x=20.\frac{1}{2}=10\\y=20.\frac{1}{4}=5\\z=20.\frac{1}{5}=4\end{cases}}\)
Vậy x = 10; y = 5 và z = 4
a)\(\frac{x}{5}=\frac{y}{6};\frac{y}{2}=\frac{z}{3}\)va \(x^3-2x^2y+z^3\)
\(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)
Ta có : | 2x - 1 | + | 1 - 2x | = 8
\(\Leftrightarrow\)| 2x - 1 | + | 2x - 1 | = 8
\(\Leftrightarrow\)2 . | 2x - 1 | = 8
\(\Leftrightarrow\)| 2x - 1 | = 4
\(\Rightarrow\)2x - 1 = 4 hoặc 2x - 1 = - 4
\(\Rightarrow\)x = 5 / 2 hoặc x = - 3 / 2
Vậy : x = 5 / 2 hoặc x = - 3 / 2