tìm x
a, x-96=(443-x)-15
b , 7x(-x-10)=0
c. 17*(3x-6)*(2x-8)=0
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`(11-x):(-5)=15`
`=> 11-x=-75`
`=> x=86`
Vậy `x = 86`
`((2x+3)^2022) . (-7x+84)=0`
`=> (2x+3)^2022 = 0` hoặc `-7x + 84 = 0`
`=> 2x+3=0` hoặc `-7x = -84`
`=> x = -3/2` hoặc `x = 12`
Vậy `x = -3/2` hoặc `x = 12`
`(x-3)(x+1) < 0`
Ta có: `x - 3 < x + 1`
nên: `x - 3 < 0` và `x + 1 > 0`
`=> x < 3 và x > -1`
`=> -1 < x < 3`
Vậy `-1 <x < 3`
#\(N\)
`a, (11-x)`\(\div\)`(-5)=15`
`11-x=15.-5`
`11-x=-75`
`x=11-(-75)`
`x=11+75=86`
`b, (2x-3)^2022.(-7x+84)=0`
`=>`\(\left\{{}\begin{matrix}\left(2x-3\right)^{2022}=0\\\left(-7x+84\right)=0\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}2x-3=0\\-7x+84=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3+0\\-7x=0-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3\\-7x=-84\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\div2\\x=-84\div-7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=12\end{matrix}\right.\)
`c, (x-3) (x+1) <0`
`-> (x-3) < x+1`
`-> (x-3)<0 , (x+1)>0`
`-> x < 3 , x> (-1)`
`-> 3 > x > -1`
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Bài 1:
a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Bài 2:
a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)
\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)
a, 7x . (x-10 ) = 0
=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)
b,17 . (3x-6).(2x-8) = 0
(3x-6).(2x-8) = 0 : 17
(3x-6).(2x-8) = 0
=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)
Nhớ nha
\(\text{a, 7x. (x-10)= 0 }\)
\(\Rightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
\(\text{ b, 17.(3x-6) . (2x-8) =0}\)
\(\left(51x-102\right).\left(2x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}51x-102=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)
a) \(7x\left(x-10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy \(x\in\left\{0;10\right\}\)
b) \(17\left(3x-6\right)\left(2x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)
Vậy \(x\in\left\{2;4\right\}\)
a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
a) -2 (x+6)+6 (x-10)=8
-2.x + (-2).6 + 6.x - 6.10 = 8
-2x + (-12) + 6x - 60 = 8
-2x + 6x + [60 + (-12)] = 8
4x + 48 = 8
4x = 8 - 48
4x = -40
x = -40 : 4
x = -10
vậy_____
b) ( x-12)-15=(20-7) - (18-x)
x - 12 - 15 = 13 - 18 + x
x - 27 = -5 + x
x - x = -5 + 27
0x = 22
x = 22 : 0
=> x ∈ ∅
c)2(5+3x)+x=31
2.5 + 2.3x + x = 31
10 + 6x + x = 31
6x + x = 31 - 10
7x = 21
x = 21 : 7
x = 3
vậy_____
d) -x (x+7) ( x-4) =0
=> -x = 0 hoặc x+ 7 = 0 hoặc x - 4 = 0
=> x = 0 hoặc x = 0 - 7 hoặc x = 0 + 4
=> x = 0 hoặc x = -7 hoặc x = 4
vậy_____
e) x-96 = (443 - x) - 15
x - 96 = 443 - x - 15
x - 96 = (443 - 15) - x
x - 96 = 428 - x
x + x = 428 + 96
2x = 524
x = 524 : 2
x = 262
vậy_____
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
tham khảo nek: https://olm.vn/hoi-dap/detail/9664557962.html
a) x - 96 = (433 - x) - 15
x - 96 = 433 - x - 15
2x = 433 - 15 + 96
2x = 514
x = 257
b) 7x (-x-10) = 0
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-x-10=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-10\end{cases}}\)
c) 17 (3x - 6) (2x - 8) = 0
(3x - 6) (2x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)