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th1: x<-2 => 2(-x-2)+4-x=11 <=> -3x=11 => x=-11/3 (t/m đk)
th2: \(-2\le x\le4\)=> 2x+4+4-x=11 <=> x=3(t/m đk)
th3: x>4 => 2x+4+x-4=11 <=> 3x=11 <=> 11/3 (t/m đk)
=> x=11/3; x=-11/3 hoặc x=3
b. Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\) (1)
\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{10}=\frac{z}{8}\)(2)
Từ (1) và (2) => \(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{x+y+z}{15+10+8}=\frac{11}{33}=\frac{1}{3}\)
\(\frac{x}{15}=\frac{1}{3}\Rightarrow x=\frac{1}{3}\cdot15=5\) \(\frac{y}{10}=\frac{1}{3}\Rightarrow y=\frac{1}{3}\cdot10=\frac{10}{3}\)
\(\frac{z}{8}=\frac{1}{3}\Rightarrow z=\frac{1}{3}\cdot8\Rightarrow z=\frac{8}{3}\)
c. Ta thấy: \(\left(x+2\right)^{n+1}\ge0,\left(x+2\right)^{n+11}\ge0\) với mọi x.
Mà \(\left(x+2\right)^{n+1}=\left(x+2\right)^{n+11}\Rightarrow x+2\in\left\{0,1,-1\right\}\)
TH1: x + 2 = 0 => x = 0 - 2 => x = -2
TH2: x + 2 = 1 => x = 1 - 2 => x = -1
TH3: x + 2 = -1 => x = -1 - 2 => x = -3
1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
\(\left\{{}\begin{matrix}A\left(x\right)-B\left(x\right)=-x^2+4x+11\\A\left(x\right)+2B\left(x\right)=x^2-x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3B\left(x\right)=-2x^2+5x+7\\A\left(x\right)-B\left(x\right)=-x^2+4x+11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B\left(x\right)=\dfrac{2}{3}x^2-\dfrac{5}{3}x-\dfrac{7}{3}\\A\left(x\right)=-x^2+4x+11+\dfrac{2}{3}x^2-\dfrac{5}{3}x-\dfrac{7}{3}=-\dfrac{1}{3}x^2+\dfrac{7}{3}x+\dfrac{26}{3}\end{matrix}\right.\)
1) =>2x+4+4+x=11
=>2x+4+4+x-11=0
=>3x-3=0
=>3x=3
=> x=1
Vậy x thuộc {1}
2)=>x+x+1+2x+4=3
=>x+x+1+2x+4-3=0
=>4x+2=0
=>4x=-2
=>x=-2/4
=>x=-1/2
Vậy x thuộc {-1/2}