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a) Ta có: \(x^2-25=0\)
\(\Leftrightarrow x^2=25\)
hay \(x\in\left\{5;-5\right\}\)
Vậy: \(x\in\left\{5;-5\right\}\)
b) Ta có: \(\left(x-4\right)^2-36=0\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=6\\x-4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{10;-2\right\}\)
c) Ta có: \(\left(6-x\right)^2=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Vậy: x=6
d) Ta có: \(\left(x+\frac{1}{4}\right)^2-\frac{1}{9}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2=\frac{1}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{1}{3}\\x+\frac{1}{4}=\frac{-1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{12}\\x=\frac{-7}{12}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{12};\frac{-7}{12}\right\}\)
\(a,96-3\left(x+8\right)=42\\ \Rightarrow3\left(x+8\right)=54\\ \Rightarrow x+8=18\\ \Rightarrow x=10.\\ b,15.5\left(x-25\right)-225=0\\ \Rightarrow75\left(x-25\right)-225=0\\ \Rightarrow75\left(x-25\right)=225\\ \Rightarrow x-25=3\\ \Rightarrow x=28.\\ c,250:x+15=25\\ \Rightarrow250:x=10\\ \Rightarrow x=25\\ d,36:\left(x-5\right)=2^2\\ \Rightarrow36:\left(x-5\right)=4\\ \Rightarrow x-5=9\\ \Rightarrow x=14.\\ e,\left[3.\left(70-x\right)+5\right]:2=46\\ \Rightarrow3.\left(70-x\right)+5=92\\ \Rightarrow3\left(70-x\right)=87\\ \Rightarrow70-x=29\\ \Rightarrow x=41.\)
1) ( 2x + 1 )2 = 25
=> ( 2x + 1 )2 = 52
=> 2x + 1 = 5 hoặc 2x + 1 = -5
=> 2x = 4 hoặc 2x = -6
=> x = 2 hoặc x = -3
2) 5x+2 = 625
=> 5x+2 = 54
=> x + 2 = 4
=> x = 2
3) ( 2x - 3 )2 = 36
=> ( 2x - 3 )2 = 62
=> 2x - 3 = 6 hoặc 2x - 3 = -6
=> 2x = 9 hoặc 2x = -3
=> x = 9/2 hoặc x = -3/2
4) ( 2x - 1 )3 = -8
=> ( 2x - 1 )3 = ( -2 )3
=> 2x - 1 = -2
=> 2x = -1
=> x = -1/2
a ) 2 - x = 17 - ( - 5 )
2 - x = 17 + 5
2 - x = 22
x = 2 - 22
x = - 20
Vậy x = - 20
1) x - 36 + 12 = - x+ 10
=> x + x = 10 + 24
=> 2x = 34
=> x = 34/2 = 17
2) (x + 15) - (11 - x) = (-2)2
=> x + 15 - 11 + x = 4
=> 2x = 4 - 4
=> 2x = 0
=> x = 0
3) 40 - 4x2 = (-6)2
=> 40 - 4x2 = 36
=> 4x2 = 40 - 36
=> 4x2 = 4
=> x2 = 1
=> x = \(\pm\)1
4) (-50) + 10x2 = (-25) x |-2|
=> -50 + 10x2 = -50
=> 10x2 = -50 + 50
=> 10x2 = 0
=> x2 = 0
=> x = 0
5) |x + 1| = 2020
=> \(\orbr{\begin{cases}x+1=2020\\x+1=-2020\end{cases}}\)
=> \(\orbr{\begin{cases}x=2019\\x=-2021\end{cases}}\)
6) (x + 1)5 + 8 = 0 (xem lại đề)
7) (-20) + x3 : 16 = -24
=> x3 : 16 = -24 + 20
=> x3 : 16 = -4
=> x3 = -4 . 16
=> x3 = -64 = (-4)3
=> x = -4
9) x14 = x17
=> x14 - x17 = 0
=> x14(1 - x3) = 0
=> \(\orbr{\begin{cases}x^{14}=0\\1-x^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
10) (-36) + (1 - x)2 = 0
=> (1 - x)2 = 36
=> (1 - x)2 = 62
=> \(\orbr{\begin{cases}1-x=6\\1-x=-6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
1. 3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75+25)
3x - 36 = 12 . 100
3x - 36 = 1200
3x = 1200 + 36
3x = 1236
=> x = 1236 : 3 = 412
câu 1 thôi nhá bạn
3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75 + 25)
3x - 36 = 1200
3x = 1164
x = 388
x : 2 - 12 = 33 . 40 + 33 . 59 + 33
x : 2 - 12 = 33 . 40 + 33 . 59 + 33 . 1
x : 2 - 12 = (40 + 59 + 1)
x : 2 - 12 = 3300
x : 2 = 3288
x = 1644
(x - 4) . (9 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=4\\x=9\end{cases}}\)
(x - 6) . (2020 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=6\\x=2020\end{cases}}\)
x . (6 - x) = 0
Thỏa mãn điều kiện 6 - x = 0
x = 6
(x - 3 - 12) . (20 - x) = 0
Thỏa mãn điều kiện \(x\le20\); 20 - x = 0
x = 20
\(e,112-45+5x=87\)
\(67+5x=87\)
\(5x=20\)
\(x=4\)
\(f,6^2+64:\left(x-1\right)=52\)
\(36+64:\left(x+1\right)=52\)
\(64:\left(x+1\right)=16\)
\(x+1=4\)
\(x=3\)