a 2x+11=3(x-9) b -15(x-2)+7(3-x)=7 Giúp mk làm NHANH 2 ý này vs, cảm ơn trc nhé
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: =>15-x=-10
hay x=25
a: =>-2x+17=9
=>-2x=-8
hay x=4
d: \(\Leftrightarrow9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
Tìm x :
a) | x + 12x | = 2x
=> \(\orbr{\begin{cases}13x=2x\\13x=-2x\end{cases}}\)
=> \(\orbr{\begin{cases}11x=0\\15x=0\end{cases}}\)
=> \(x=0\)
b) 3x − |x + 1| = 1
=> |x + 1| = 3x -1
=>\(\orbr{\begin{cases}x+1=3x-1\\x+1=1-3x\end{cases}}\)
=> \(\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
c) |2x + 3| = x + 1
=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)
b) 3x - |x + 1| = 1
<=> |x + 1| = 3x - 1 (1)
ĐK : \(x\ge\frac{1}{3}\)
Khi đó (1) <=> \(\orbr{\begin{cases}x+1=3x-1\\x+1=-3x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{loại}\right)\\x=1\end{cases}}\)
Vậy x = 1
c) ĐK : x + 1\(\ge0\Rightarrow x\ge-1\)
Khi đó |2x + 3| = x + 1
<=> \(\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\left(\text{loại}\right)\)
Vậy \(x\in\varnothing\)
d) ||x + 9| + 11| = 2x + 11 (1)
ĐK : \(2x+11\ge0\Rightarrow x\ge-\frac{5}{2}\)
Khi đó (1) <=> \(\orbr{\begin{cases}\left|x+9\right|+11=2x+11\\\left|x+9\right|+11=-2x-11\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left|x+9\right|=2x\\\left|x+9\right|=-2x-22\end{cases}}\)
Khi |x + 9| = 2x (x \(\ge0\))
<=> \(\orbr{\begin{cases}x+9=2x\\x+9=-2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\left(tm\right)\\x=-3\left(\text{loại}\right)\end{cases}}\)
Khi |x + 9| = -2x - 22 ( \(-\frac{5}{2}\le x\le-11\))
<=> \(\orbr{\begin{cases}x+9=-2x-22\\x+9=2x+22\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{31}{3}\\x=-13\end{cases}}\left(\text{loại}\right)}\)
Vậy x = 9
a: =>-2x=9-17=-8
hay x=4
b: =>15-x=-10
hay x=25
d: \(\Leftrightarrow9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
e: \(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\3-x=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
a: \(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>13/12x=13/12
hay x=1
b: \(\Leftrightarrow\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)
\(\Leftrightarrow\dfrac{3}{11}x-1-\dfrac{1}{3}x=\dfrac{3}{7}x-\dfrac{5}{7}-\dfrac{5}{9}x+\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{46}{693}=\dfrac{13}{21}\)
hay x=429/46
a: =>-2x+17=9
=>-2x=-8
hay x=4
b: =>15-x=-10
hay x=25
c: =>7x=-4
hay x=-4/7
d: =>\(9x^2=81\)
hay \(x\in\left\{3;-3\right\}\)
g: \(\Leftrightarrow2x-4+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)
=>\(\left(x-2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)
nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
=>\(2^x\left(1+2+2^2+2^3\right)=120\)
=>\(2^x\cdot15=120\)
=>\(2^x=8\)
=>x=3
e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
a, 2x+11=3(x-9) b,-12(x-2)+7(3-x)=7
2x+11=3x-27 45-19x =7
-x=-38 19x =38
x=38 x =2
-12x+60+7x-21=5
-5x+39=5
-5x =-34
x =6.8