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SG
2
S
27 tháng 10 2015
a/ => (x2 - 5)(x + 5)(x - 5) = 0
=> x2 - 5 = 0 => x2 = 5 => x = \(+-\sqrt{5}\) (loại)
hoặc x + 5 = 0 => x = -5
hoặc x - 5 = 0 => x = 5
Vậy x = 5 ; x = -5
b/ => x + 5 = 0 => x = -5
hoặc 9 + x2 = 0 => x2 = -9 (vô nghiệm)
Vậy x = -5
c/ => x + 3 = 0 => x = -3
hoặc x2 + 1 = 0 => x2 = -1 (vô nghiệm)
Vậy x = -3
d/ => (x + 5)(x + 2)(x - 2) = 0
=> x + 5 = 0 => x = -5
hoặc x + 2 = 0 => x = -2
hoặc x - 2 = 0 => x = 2
Vậy x = -5 ; x = -2; x = 2
HB
23 tháng 1 2018
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
11 tháng 1 2018
a)
\(\left|x\right|-2\left|x\right|+3\left|x\right|=16+6\left|x\right|-19\)
\(\left|x\right|-2\left|x\right|+3\left|x\right|-6\left|x\right|=16-19\)
\(\left|x\right|.\left(1-2+3-6\right)=-3\)
\(\left|x\right|.\left(-4\right)=-3\)
\(\left|x\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
b,
2.(|x| - 5) - 15 = 9
\(2.\left(\left|x\right|-5\right)=9+15\)
\(2.\left(\left|x\right|-5\right)=24\)
\(\left|x\right|-5=24:2\)
\(\left|x\right|-5=12\)
\(\left|x\right|=12+5\)
\(\left|x\right|=17\)
\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
c,
|8 - 2x| + |4y - 16| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|8-2x\right|=0\\\left|4y-16\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}8-2x=0\\4y-16=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=8\\4y=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
d,
|x - 14| + |2y - x| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|x-14\right|=0\\\left|2y-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-14=0\\2y-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
2.Tìm x, y, z biết
a,
2.|3x| + |y + 3| + |z - y| = 0
\(\Rightarrow\left\{{}\begin{matrix}2.\left|3x\right|=0\\\left|y+3\right|=0\\\left|z-y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x\right|=0\\y+3=0\\z-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\y=-3\\z=y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
b, (x - 3y)2 + | y + 4|= 0
\(\Rightarrow\left\{{}\begin{matrix}\left(x-3y\right)2=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-4\right)\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
AH
11 tháng 7 2018
1.
a) (—7/3)3:(—7/3)2=(—7/3)3–2=—7/3
b) (—4/9):(—4/9)3= (—4/9)1–3=(—4/9)—2=81/16
c) (1/5)10:(1/5)7=(1/5)10–7=(1/5)3=1/125
2.
a) —x/7 =1/—21
==> —x.(—21)=7.1
==> —x.(—21)=7
==> —x=7:(—21)
==> —x=—1/3
==> x=1/3
b) 4 2/5 . 0,5–1 3/7= 22/5 . 1/2 —10/7= 22.1/5.2–10/7= 11/5 —10/7= 77/35 — 50/35= 27/35
c) 3x2–2x=0
==> x3(3–2)=0
x3.1=0
x3=0:1
x3=0
==> x=0
c) 9x2–1=0
9x2=0+1
9x2=1
x2=1:9
x2=1/9
x2=12/32 hoặc x2=(—1/3)2
Vậy x=1/3 hoặc x=—1/3
21 tháng 7 2022
Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
25 tháng 6 2019
\(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)
\(\Rightarrow-7x=\frac{-1}{6}\)
\(\Rightarrow x=\frac{1}{42}\)
Vậy ...
\(\)
25 tháng 6 2019
\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{18}\)
Vậy...
\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)
\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)
Vậy ...
\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)
\(\text{Vì }9x^2\ge0\)
\(\Rightarrow9x^2+9\ge9\)
\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x.\left(x^2+1\right)+9.\left(x^2+1\right)=0\\x.\left(x^2+1\right)-9.\left(x^2+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)