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1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+8+2x+4\right)\left(3x+8-2x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(5x+12\right)\left(x+4\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: \(\Leftrightarrow\left|4x+2\right|=x+15\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(4x+2+x+15\right)\left(4x+2-x-15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(5x+17\right)\left(3x-13\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{17}{5};\dfrac{13}{3}\right\}\)
c: =>3x+7>=0
hay x>=-7/3
d: =>|2x-5|=-2x+5
=>2x-5<=0
hay x<=5/2
a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)
Bài 3: Tìm x:
a. \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
=> 2x - 1 = 3
=> 2x = 4
=> x = 2
b. \(\left(x-2\right)^2=1\)
\(\Rightarrow\) \(\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c. \(x^{2000}=x\)
=> x = 1
d. \(\left(4x-3\right)^3=-125\)
\(\Rightarrow\left(4x-3\right)^3=\left(-5\right)^3\)
=> 4x - 3 = -5
=> 4x = -2
=> x = \(\dfrac{-1}{2}\)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-15=0\\2x-15=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=15\\2x=16\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{15}{2}\\x=8\end{array}\right.\)
a) \(\left(2x-1\right)^{10}=\left(1-2x\right)^5\)
\(\Rightarrow\left(2x-1\right)^2=1-2x\)
\(\Rightarrow4x^2-4x+1=1-2x\)
\(\Rightarrow4x^2-4x=-2x\)
\(\Rightarrow2x^2-2x=-x\)
\(\Rightarrow2x^2-x=0\)
\(\Rightarrow x.\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
b) \(\left(3x-1\right)^{15}=\left(1-3x\right)^8\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(1-3x\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(-\left(3x-1\right)\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(3x-1\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^8.\left(\left(3x-1\right)^7-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^8=0\\\left(3x-1\right)^7-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....
c) Tự lm
a, \(\left|x+1\right|-4=0\)
\(\Rightarrow\left|x+1\right|=4\)
+) Xét \(x\ge-1\) có:
\(x+1=4\Rightarrow x=3\) ( t/m )
+) Xét x < -1 có:
\(-x-1=4\Rightarrow x=-5\) ( t/m )
Vậy x = 3 hoặc x = -5
b, tương tự
c, \(3x+\left|2x\right|=5x\)
\(\Rightarrow\left|2x\right|=2x\)
+) Xét \(x\ge0\)
\(\Rightarrow2x=2x\Rightarrow x\in R\forall x>0\)
+) Xét x < 0
\(\Rightarrow-2x=2x\Rightarrow x=0\)
Vậy \(x\ge0\)
d, \(\left|2x+1\right|=\left|x-3\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=x-3\\2x+1=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...
a) Đặt 2 x - 15 = t
TA có :
\(t^5=t^3\) => \(t^5-t^3=0\Leftrightarrow t^3\left(t^2-1\right)=0\)
=> t^3 = 0 hoặc t^2 - 1 = 0
=> t =0 hoặc t^2 = 1
=> t = 0 hoặc t = 1 hoặc t = -1
(+) t = 0 => 2x - 15 = 0 => x = 15/2
(+) 2x- 15 = 1 => 2x = 16 => x = 8
(+) 2x- 1 5 = -1 => 2x = 14 => x = 7
b) x^2 < 5
=> x < \(\sqrt{5}\approx2,2\)
Vì x thuộc N => x = { 0;1;2)
a) (2x-15)5 = (2x - 15)3
=> 2x - 15 = 1; 2x - 15 = - 1 ; 2x - 15 = 0
TH1: 2x - 15 = 1
=> 2x = 15 + 1= 16 (chọn vì là STN)
x = 16 : 2 = 8
TH2: 2x - 15 = - 1
2x = -1 + 15 = 14
=> x = 14 : 2 = 7 (chọn vì là STN)
TH2: 2x - 15 = 0
2x = 0 + 15 = 15
=> x = 15: 2 = 7,5 (loai vì là số thập phân)
=> x = 7 ; hoặc x = 8