A) (x-2)^2=1/16
B) (x-1)^2=(1-x)^4
C) (3x-1/5)^2=(2/3)^6
Đ) 3x/27=3^-1
K
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AK
15 tháng 4 2018
\(a)\frac{5}{8}-x=2\frac{1}{6}\)
\(\Rightarrow\frac{5}{8}-x=\frac{13}{6}\)
\(\Rightarrow x=\frac{5}{8}-\frac{13}{6}\)
\(\Rightarrow x=\frac{15}{24}-\frac{52}{24}\)
\(\Rightarrow x=-\frac{37}{24}\)
\(b)\) \(\frac{4}{9}:x=-\frac{1}{3}+1\frac{1}{6}\)
\(\Rightarrow\frac{4}{9}:x=-\frac{1}{3}+\frac{7}{6}\)
\(\Rightarrow\frac{4}{9}:x=-\frac{2}{6}+\frac{7}{6}\)
\(\Rightarrow\frac{4}{9}:x=\frac{5}{6}\)
\(\Rightarrow x=\frac{4}{9}:\frac{5}{6}\)
\(\Rightarrow x=\frac{4}{9}.\frac{6}{5}\)
\(\Rightarrow x=\frac{8}{15}\)
\(c)\left(3x-2\right)^3=-\frac{1}{27}\)
\(\Rightarrow3x-2=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+2\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{6}{3}\)
\(\Rightarrow3x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:3\)
\(\Rightarrow x=\frac{5}{9}\)
d ) và e ) tự làm
Chúc bạn học tốt !!!
15 tháng 4 2018
5/8-x=21/6
x=5/8-21/6
x=-23/8
Vậy x=-23/8
b, 4/9:x=-1/3+11/6
4/9:x= 3/2
x= 4/9:3/2
20 tháng 3 2020
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
20 tháng 3 2020
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
RA
a) (2x-1)^3=27
b) (2x-1)^4=81
c) (x-2)^5=-32
d) (3x-1)^4=(3x-1)^6
đ) 5^x +5^x+2=650
g) 3^x-1 +5.3^x-1=162
6
26 tháng 8 2019
a) (2x-1)3 = 27
(2x-1)3 = 93
2x-1 = 9
2x = 9+1
2x = 10
x = 10:5
x = 2
Vậy x = 2
26 tháng 8 2019
b) (2x-1)4 = 81
(2x-1)4 = (\(\pm\)34)
2x-1 = \(\pm\)3
Trường hợp 1:
2x-1 = 3
2x = 3+1
2x = 4
x = 4:2
x = 2
Trường hợp 2:
2x-1 = -3
2x = -3+1
2x = -2
x = -2:2
x = -1
Vậy x \(\in[_{ }2;-1]\)
Vì không tìm thấy ngoặc nhọn nên mình dùng tạm ngoặc vuông nhé
HY
20 tháng 3 2020
\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
HY
20 tháng 3 2020
\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình trên là \(-2\)
14 tháng 3 2020
a) ĐKXĐ: \(x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> \(5+\frac{96}{\left(x-4\right)\left(x+4\right)}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
<=> 5(x - 4)(x + 4) + 96(x - 4) = (2x - 1)(x - 4)(4 - x) - (3x - 1)(x + 4)(4 - x)
<=> 20x2 - 16x + 64 = 18x2 + 8x
<=> 20x2 - 16x + 64 - 18x2 - 8x = 0
<=> 2x2 - 24x + 64 = 0
<=> 2(x2 - 12x + 32) = 0
<=> 2(x - 8)(x - 4) = 0
<=> (x - 8)(x - 4) = 0
<=> x - 8 = 0 hoặc x - 4 = 0
<=> x = 8 (tm) hoặc x - 4 = 0 (ktm)
=> x = 8
b) ĐKXĐ: \(x\ne\pm\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-2^2}\)
<=> \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
<=> (2 + 3x)2 - 6(3x - 2) = 9x2
<=> 16 - 6x + 9x2 = 9x2
<=> 16 - 6x + 9x2 - 9x2 = 0
<=> 16 - 6x = 0
<=> -6x = 0 - 16
<=> -6x = -16
<=> x = -16/-6 = 8/3
=> x = 8/3
1 tháng 1 2018
a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)
b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)
\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x^3+1\right)}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{x-1}\cdot\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{-\left(x-1\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{-2\left(1+x+x^2\right)}{2x+3y}\)
HN
1 tháng 1 2018
a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)
\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)
\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)
b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4x^2-16.\dfrac{7x-2}{3x+6}\)
\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3x^3+3}{x-1}.\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x^3+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)
\(=\dfrac{2\left(2x+3y\right)\left(1-x\right)\left(1+x+x^2\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(2x+3y\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(x^2+x+1\right)}{2x+3y}\)
20 tháng 10 2018
1
a) \(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(x+5y\right)\left(x-5y\right)=x^2-25y\)
b) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)
\(\left(x-5\right)\left(x^2+5x+25\right)=x^3-125\)
26 tháng 10 2022
Bài 3:
a: \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
=>8x+1=0
=>x=-1/8
b: \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
=>2x+255=0
=>x=-255/2
c: \(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6=49\)
=>24x+62=49
=>24x=-13
=>x=-13/24
d: =>x^3+8-x^3-2x=15
=>-2x=15-8=7
=>x=-7/2
A) \(\left(x-2\right)^2=\dfrac{1}{16}\\ Mà:\left(\dfrac{1}{4}\right)^2=\dfrac{1}{16}hoặc\left(-\dfrac{1}{4}\right)^2=16\\ =>\left(x-2\right)^2=\left(\dfrac{1}{4}\right)^2hoặc\left(x-2\right)^2=\left(-\dfrac{1}{4}\right)^2\\ =>x-2=\dfrac{1}{4}hoặc\left(x-2\right)=-\dfrac{1}{4}\\ =>\left[{}\begin{matrix}x=\dfrac{1}{4}+2\\x=-\dfrac{1}{4}+2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\)