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a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
\(e,112-45+5x=87\)
\(67+5x=87\)
\(5x=20\)
\(x=4\)
\(f,6^2+64:\left(x-1\right)=52\)
\(36+64:\left(x+1\right)=52\)
\(64:\left(x+1\right)=16\)
\(x+1=4\)
\(x=3\)
CÂU 10:
a, -x - 84 + 214 = -16 b, 2x -15 = 40 - ( 3x +10 )
x = - ( -16 -214 + 84 ) 2x + 3x = 40 -10 +15
x = 16 + 214 - 84 5x = 45
x = 146 x = 9
c, \(|-x-2|-5=3\) d, ( x - 2)(2x + 1) = 0
\(|-x-2|=8\) => x - 2 = 0 hoặc 2x + 1 = 0
=> - x - 2 = 8 hoặc x + 2 = 8 \(\orbr{\begin{cases}x-2=0\\2x+1=0\end{cases}=>}\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
\(\orbr{\begin{cases}-x-2=8\\x+2=8\end{cases}=>\orbr{\begin{cases}x=-10\\x=6\end{cases}}}\)
Bài 1 :Tính :
A = 2^4 . 417 + (-2)^4 . 583
B = 146.572 + (-146).(-428)
C = (- 158 ) . 1999 + 842 . (-1999)
D = 76 - 2x + 24 - 2y với x + y = - 50
Bài 2 . tìm x
a) /2x-6/ - / x - 12 / = 0
b)/ x + 5 / + ( y - 3 ) ^ 2 = 0
Bài 3 Tìm x
a) 4x - 11 = - 6x + 89
b) ( 3x - 5 ) - ( 2x -7 )=-16
c) / 2x - 4 / + 11 = 19
d) ( x - 3 ) ^ 2 - 25 = 0
nhiều bài quá mk làm ko nổi
xin lỗi bn nha!
Vũ Vân Anh 
shi nit chi
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
Bài 2:
a: (2x-1)(2y+1)=35
=>\(\left(2x-1;2y+1\right)\in\left\{\left(1;35\right);\left(35;1\right);\left(-1;-35\right);\left(-35;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(1;17\right);\left(18;0\right);\left(0;-18\right);\left(-17;-1\right)\right\}\)
b: (5x+2)(y-3)=14
=>\(\left(5x+2;y-3\right)\in\){(1;14);(14;1);(-1;-14);(-14;-1);(2;7);(7;2);(-2;-7);(-7;-2)}
=>(x;y)\(\in\left\{\left(-\dfrac{1}{5};17\right);\left(\dfrac{12}{5};4\right);\left(-\dfrac{3}{5};-11\right);\left(-\dfrac{16}{5};2\right);\left(0;10\right);\left(1;5\right);\left(-\dfrac{4}{5};-4\right);\left(-\dfrac{9}{5};1\right)\right\}\)
mà x,y nguyên
nên \(\left(x;y\right)\in\left\{\left(0;10\right);\left(1;5\right)\right\}\)
c: y-6x+2xy=10
=>2xy-6x+y=10
=>2x(y-3)+y-3=7
=>(2x+1)(y-3)=7
=>\(\left(2x+1;y-3\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;10\right);\left(2;4\right);\left(-1;-4\right);\left(-4;2\right)\right\}\)
Bài 1:
a: \(15-3\left(2x-1\right)=-12\)
=>3(2x-1)=15+12=27
=>2x-1=9
=>2x=10
=>x=5
b: \(4\left(3x+2\right)-17=27\)
=>4(3x+2)=27+17=44
=>3x+2=11
=>3x=9
=>x=3
c: \(18-3\left(2x+1\right)^2=-57\cdot2\)
=>\(3\left(2x+1\right)^2=18+57\cdot2=132\)
=>\(\left(2x+1\right)^2=\dfrac{132}{3}=44\)
=>\(\left[{}\begin{matrix}2x+1=2\sqrt{11}\\2x+1=-2\sqrt{11}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{11}-1}{2}\left(loại\right)\\x=\dfrac{-2\sqrt{11}-1}{2}\left(loại\right)\end{matrix}\right.\)
d: \(\left(3x-2\right)\cdot3-7\cdot\left(-8\right)=120\)
=>\(3\left(3x-2\right)+56=120\)
=>3(3x-2)=120-56=120-20-36=100-36=64
=>\(3x-2=\dfrac{64}{3}\)
=>\(3x=\dfrac{64}{3}+2=\dfrac{70}{3}\)
=>\(x=\dfrac{70}{9}\)
e: \(\left(25-x^2\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}25-x^2=0\\x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x^2=25\\x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\\x=-5\\x=-3\end{matrix}\right.\)
f: Sửa đề: \(\left(x+3\right)^5=\left(x+3\right)^3\)
=>\(\left(x+3\right)^5-\left(x+3\right)^3=0\)
=>\(\left(x+3\right)^3\cdot\left[\left(x+3\right)^2-1\right]=0\)
=>\(\left(x+3\right)\left(x+3+1\right)\left(x+3-1\right)=0\)
=>(x+3)(x+4)(x+2)=0
=>\(\left[{}\begin{matrix}x+3=0\\x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\\x=-2\end{matrix}\right.\)
g: Sửa đề: \(\left(2x-1\right)^7=27\left(2x-1\right)^4\)
=>\(\left(2x-1\right)^7-27\cdot\left(2x-1\right)^4=0\)
=>\(\left(2x-1\right)^4\cdot\left[\left(2x-1\right)^3-1\right]=0\)
=>\(\left[{}\begin{matrix}\left(2x-1\right)^4=0\\\left(2x-1\right)^3-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
a; 15 - 3.(2\(x\) - 1) = - 12
3.(2\(x-1\)) = 15 - (-12)
3.(2\(x\) - 1) = 27
2\(x-1\) = 27 : 3
2\(x\) - 1 = 9
2\(x\) = 9 + 1
2\(x\) = 10
\(x=10:2\)
\(x=5\)
Vậy \(x=5\)
b; 4.(3\(x+2\)) - 17 = 27
4.(3\(x\) + 2) = 27 + 17
4.(3\(x\) + 2) = 44
3\(x\) + 2 = 44 : 4
3\(x\) + 2 = 11
3\(x\) = 11 - 2
3\(x\) = 9
\(x\) = 9 : 3
\(x=3\)
Vậy \(x=3\)