Tìm x, biết:
\(\left|2x+1\right|=4-x\)
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mọi người ơi câu b là giá trị tuyệt đối của x^2 -1 nha
giúp mình mình tick cho
a) \(\Leftrightarrow x^2+\dfrac{2}{3}x-x^2+\dfrac{3}{4}x=\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{17}{12}x=\dfrac{7}{12}\Leftrightarrow x=\dfrac{7}{17}\)
c) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=-1\\2x+1=1\\2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left|\left(x+\frac{1}{2}\right).\left|2x-\frac{3}{4}\right|\right|=2x-\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|.\left|2x-\frac{3}{4}\right|=2x-\frac{3}{4}\)
\(\Rightarrow2x-\frac{3}{4}\ge0\) (1)
Lúc này ta có: \(\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)=2x-\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)-\left(2x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left(2x-\frac{3}{4}\right).\left(\left|x+\frac{1}{2}\right|-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=0\\\left|x+\frac{1}{2}\right|-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{3}{4}\\\left|x+\frac{1}{2}\right|=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x+\frac{1}{2}=1\\x+\frac{1}{2}=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x=\frac{1}{2}\\x=\frac{-3}{2}\end{array}\right.\)
Mà \(x\ge\frac{3}{8}\) do \(2x-\frac{3}{4}\ge0\)
Vậy \(x\in\left\{\frac{3}{8};\frac{1}{2}\right\}\)
\(49\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)
\(4x+2=0\)
\(4x=2\)
\(x=-\frac{1}{2}\)
<=>4(x2+2x+1)+4x2-4x+1-8x2+8-11=0
<=>4x2+8x+4+4x2-4x+1-8x2+8-11=0
<=>4x+2=0
<=>2(2x+1)=0
<=>2x+1=0
<=>x=-1/2
b) 3x - 6 - (8x + 4) - (10x + 15) = 50
=> 3x - 6 - 8x - 4 - 10x - 15 = 50
=> (3x - 8x - 10x) = 6+ 4 + 15 + 50
=> -15x = 75 => x = 75 : (-15) = -5
c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)
+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3
+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1
Vậy x = 5/3 hoặc x = 1
a) (n-1)n+11-(n-1)n=0
(n-1)n(n-1)11-(n-1)n=0
(n-1)n[(n-1)11-1]=0
(n-1)n=0 hoặc (n-1)11-1=0
n-1=0 hoặc (n-1)11 =1
n=1 hoặc n-1 =1
n=1 hoặc n =2
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(\left(x-1\right)^3+\left(x-3\right)^3=\left(2x-4\right)^3\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3-\left(2x-4\right)^3=0\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3+\left(4-2x\right)^3=0\)
Đặt \(\left(x-1\right)=a;\left(x-3\right)=b;\left(4-2x\right)=c\)ta có:
\(a^3+b^3+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Thay lại, ta được:
\(\left(a+b+c\right)^3=\left(x-1+x-3+4-2x\right)^3=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(x-1+x-3\right)\left(x-3+4-2x\right)\left(4-2x+x-1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(1-x\right)\left(3-x\right)=0\)
\(\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Rightarrow x=2\)
hoặc \(1-x=0\Leftrightarrow x=1\)
hay \(3-x=0\Leftrightarrow x=3\)
Vậy \(x\in\left\{1;2;3\right\}\). Xong :))
/ 2x + 1 / = 4 - x
TH1 : \(2x+1\ge0=>x\ge\frac{-1}{2}\)
PT trở thành :
\(2x+1=4-x\)
\(=>3x=3=>x=1\)thỏa mãn
TH2 : \(2x+1< 0=>x< \frac{-1}{2}\)
PT trở thành :
\(-2x-1=4-x\)
\(=>-x=5\)
\(=>x=-5\)thỏa
Tự kl nha
\(\left|2x+1\right|=4-x\left(1\right)\)
Xét 2 TH sau:
\(\left(+\right)x\ge-\frac{1}{2}\) ,khi đó (1) trở thành: \(2x+1=4-x=>2x-\left(-x\right)=4-1=>3x=3=>x=1\)
\(\left(+\right)x< -\frac{1}{2}\),khi đó (1) trở thành: \(-\left(2x+1\right)=4-x=>-2x-1=4-x=>-2x-\left(-x\right)=4-\left(-1\right)=>-x=5=>x=-5\)
Vậy x=-5;x=1