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a: =>|x+2|=2x-6
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-6-x-2\right)\left(2x-6+x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(x-8\right)\left(3x-4\right)=0\end{matrix}\right.\Leftrightarrow x=8\)
b: =>1,2x+1,3x=2,95
=>2,5x=2,95
hay x=1,18
a, |x + 2| + 6 = 2x
\(\Rightarrow\left|x+2\right|=2x-6\) (1)
Vì |x + 2| \(\ge\) 0 \(\Rightarrow2x-6\ge0\Rightarrow2x\ge6\Rightarrow x\ge3\)
Từ (1) \(\Rightarrow\left[{}\begin{matrix}x+2=2x-6\\x+2=-\left(2x-6\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2x=-6-2\\x+\left(-2x\right)=6-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\-x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=-4\end{matrix}\right.\)(Loại)
Vậy không có giá trị x thỏa mãn
a, -19 - x = -20
x = -19 - (-20)
x = -19 + 20
x = 1
b, 5x - 6 = 3x + 12
5x - 6 - 3x = 12
5x - 3x = 12 + 6
(5 - 3)x = 18
2x = 18
x = 18 : 2
x = 9
c, 15 - 3 (x - 1) = 8 - 2x
15 - 3 (x - 1) + 2x = 8
-3x - 3 - 2x = 8 - 15
-3x - 3 - 2x = -7
-3x - 2x - 3 = 7
-3x - 2x = 7 + 3
(-3 - 2) x = 10
-5x = 10
x = 10 : (-5)
x = -2
d, (5x - 6)2 = 16
(5x - 6)2 = 42
=> 5x - 6 = 4
5x = 4 + 6
5x = 10
x = 10 : 5
x = 2
f, 26 - | x + 9 | = 13
| x + 9 | = 26 - 13
=> | x + 9 | = 13
=> x + 9 = +- 13
* Với x + 9 = 13
x = 13 - 9
x = 4
* Với x + 9 = -13
x = -13 - 9
x = -22
Vậy x = {4;-22}
e, | 3 + x | = 19
=> 3 + x = +- 19
* Với 3 + x = 19
x = 19 - 3
x = 16
* Với 3 + x = -19
x = -19 - 3
x = -22
Vậy x = {16;-22}
a, X = -19+20=1
b, (5-3)X = 18
2X = 18
=> X = 9
c, 3X + 3 -2X = 7
X+3 =7
X = 4
f, |X+9| = 13
ta có 2 trường hợp:
TH1: X+9 = 13
=> X= 4
TH2 : X+9 = -13
=> X= -22
e, ta có 2 trường hợp:
TH1: 3+X = 19
=> X= 16
TH2: 3+X = -19
=> X= -22
\(x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x.\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy ...
mấy câu khác tương tự
5.\(\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
vậy.....
câu 6 tương tự câu 5
\(5.\left(x-5\right)^6=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
X : 6 x 7,2 + X x 1,3 + X : 2 + 15 = 19,95
X x \(\frac{1}{6}\) x \(\frac{36}{5}\)+ X x \(\frac{13}{10}\)+ X x \(\frac{1}{2}\) = 19,95 - 15 = 4,95 = \(\frac{99}{20}\)
X x \(\frac{6}{5}\)+ X x \(\frac{13}{10}\)+ X x \(\frac{1}{2}\)= \(\frac{99}{20}\)
X x ( \(\frac{6}{5}+\frac{13}{10}+\frac{1}{2}\)) = \(\frac{99}{20}\)
X x 3 = \(\frac{99}{20}\)
X = \(\frac{99}{20}:3\)
X = \(\frac{33}{20}\)
X : 6 x 7,2 + 1,3 x X + X : 2 + 15 = 19,95
(X x 7,2) / 6 + (7,8 x X) / 6 + (X x 3) / 6 + 90/6 = 119,7/6
X x 7,2 + 7,8 x X + X x 3 + 90 = 119,7
X x (7,2 + 7,8 + 3) = 119,7 - 90
X x 18 = 29,7
X = 29,7 : 18
X = 1,65