Bài 2.Tìm x
a,x+8=2
b,-15+4x=5
K
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26 tháng 8 2021
`a)sqrt{1-4x+4x^2}+5=x-2`
`<=>\sqrt{(2x-1)^2}=x-2-5`
`<=>|2x-1|=x-7(x>=7)`
`<=>[(2x-1=x-7),(2x-1=7-x):}`
`<=>[(x=-6(ktm)),(3x=8):}`
`<=>x=8/3(ktm)`
Vậy PTVN
`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`
`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`
`<=>4sqrt{x+3}=4`
`<=>sqrt{x+3}=1<=>x+3=1`
`<=>x=-2(tm)`
Vậy `S={-2}`
26 tháng 8 2021
a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)
TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)
Vậy \(S=\varnothing\)
b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)
12 tháng 12 2023
Bài 5:
a: x(x-4)=0
=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b: Đề thiếu vế phải rồi bạn
Bài 6:
a: \(\left(-5\right)\cdot\left(-6\right)\cdot\left(-4\right)\cdot2\)
\(=-\left(2\cdot5\right)\cdot\left(4\cdot6\right)\)
\(=-24\cdot10=-240\)
b: \(\left(-3\right)\cdot2\cdot\left(-8\right)\cdot5\)
\(=3\cdot2\cdot8\cdot5\)
\(=\left(3\cdot8\right)\cdot\left(2\cdot5\right)\)
\(=24\cdot10=240\)
LT
1
T
22 tháng 12 2023
a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
AN
3
K
20 tháng 3 2022
\(a,x=\dfrac{1}{2}-\dfrac{2}{5}\)
\(x=\dfrac{1}{10}\)
\(b,x+\dfrac{3}{7}=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}-\dfrac{3}{7}\)
\(x=\dfrac{19}{70}\)
\(c,19-x=\dfrac{17}{20}\)
\(x=19-\dfrac{17}{20}\)
\(x=\dfrac{363}{20}\)
9 tháng 11 2021
Bài 1:
\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)
Bài 2:
\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)
P
1
NN
29 tháng 12 2022
a.\(\left(-12\right)x-14=-2\)
\(\left(-12\right)x=-2+14\)
\(\left(-12\right)x=12\)
\(x=12:\left(-12\right)\)
\(x=-1\)
\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)
\(\left(-8\right)x=35-3\)
\(\left(-8\right)x=32\)
\(x=32:\left(-8\right)\)
\(x=-4\)
\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)
\(\left(-9\right)x+3=14+16\)
\(\left(-9\right)x+3=30\)
\(\left(-9\right)x=30-3\)
\(\left(-9\right)x=27\)
\(x=27:\left(-9\right)\)
\(x=-3\)
12 tháng 5 2023
a: x=5:(-1/2)=-10
b: x=8/3+1/9=25/9
c: =>x+5/6=11/21
=>x=-13/42
d: =>7/4x-5=-10/3
=>7/4x=5/3
=>x=20/21
e: =>10/3-3/4:x=-1/6
=>3/4:x=10/3+1/6=21/6=7/2
=>x=3/4:7/2=3/4*2/7=6/28=3/14
g: =>3/(x+5)=3/20
=>x+5=20
=>x=15
h: =>1-1/2+1/2-1/3+...+1/x-1/x+1=49/50
=>1-1/x+1=49/50
=>x+1=50
=>x=49
a)x=2-8
x=-6
b)4x=20
x=5
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