tim x:
a)[2x+3/5]2-9/25=0 b)23/x-5=-47/(-2)x+7
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l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)
=>7-x+5=15-x+x-12
=>12-x=3
hay x=9
b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)
\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)
=>x-38-x=13-40-x
=>-27-x=-38
=>x+27=38
hay x=11
e: \(x^2+3x+9⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)
hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
1) (3x + 9)(3x - 6) = 0
=> \(\orbr{\begin{cases}3x+9=0\\3x-6=0\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-9\\3x=6\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy ...
b) (2x + 15) - 25 = 47 - (10 - x)
=> 2x - 10 = 37 + x
=> 2x - x = 37 + 10
=> x = 47
3, tương tự
4) |4 - 3x| = 8
=> \(\orbr{\begin{cases}4-3x=8\\4-3x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-4\\3x=12\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{4}{3}\\x=4\end{cases}}\)
Vì x là số nguyên nên ...
còn lại tương tự
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: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
Bài 1:
a) Ta có: \(-5x+32=\left(-2\right)^3\)
\(\Leftrightarrow-5x+32=-8\)
\(\Leftrightarrow-5x=-40\)
hay x=8
Vậy: x=8
b) Ta có: \(-2x+36=6\)
\(\Leftrightarrow-2x=6-36=-30\)
hay x=15
Vậy: x=15
e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
b,-2x+36=6
tương đương -2x=-30
tương đương x=15
a, -5x+32=(-2)^3
tương đương -5x+32=8
tương đương -5x=-24
tương đương x=24/5
a) <=> 15-5x-20=-12-3
<=> -5x=-12-3-15+20=-10
=>x=-10:(-5)=2
b)<=>7-x-25-7=-25
<=> -x=-25-7+25+7=0 =>x=0
c) /x+2/=0 => x+2=0 =>x=-2
d) sai đề
e)<=> /x-5/ = 7
<=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)
<=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)
g) <=> -x-20-8+2x=-15
<=> x=-15+20+8=13
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