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TN
1
HN
2
MN
28 tháng 5 2021
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
7P
3
TC
7 tháng 5 2022
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
23 tháng 9 2021
d: Ta có: (-4,6)+x=-3,5
nên x=-3,5+4,6
hay x=1,1
a: Ta có: \(1.5-\left|x-0.3\right|=0\)
\(\Leftrightarrow\left|x-0.3\right|=1.5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0.3=1.5\\x-0.3=-1.5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.2\end{matrix}\right.\)
GH
2
4 tháng 9 2021
a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
AH
Akai Haruma
Giáo viên
4 tháng 9 2021
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
NT
4
PT
Phạm Trần Hoàng Anh
CTVHS
25 tháng 6 2021
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
25 tháng 6 2021
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
ND
1
26 tháng 11 2021
a, ĐKXĐ:\(x\ge1\)
\(\sqrt{x-1}=3\\ \Rightarrow x-1=9\\ \Rightarrow x=10\)
\(b,x^2-64=0\\ \Rightarrow\left(x-8\right)\left(x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\\ c,x^2+16=25\\ \Rightarrow x^2=9\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ d,ĐKXĐ:x\ge0\\ \left|\sqrt{x}-3\right|+3=9\\ \Rightarrow\left|\sqrt{x}-3\right|=6\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}-3=-6\\x-3=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vô.lí\right)\\x=9\left(tm\right)\end{matrix}\right.\)
23 tháng 11 2021
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
a) \(\left(x-3\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{4}{5}\end{array}\right.\)
b) \(\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=5\\x-1=-5\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-4\end{array}\right.\)
a)(x-3)(4-5x)\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{4}{5}\end{array}\right.\)
Vậy x=3 và \(\frac{4}{5}\)
b) \(\left(x-1\right)^2=25\Rightarrow\begin{cases}x-1=5\\x-1=-5\end{cases}\)
\(\Rightarrow\begin{cases}x=6\\x=-4\end{cases}\)
Vậy x=-4 và 6