Tìm x thuộc Q, biết rằng:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
b) \(\left(x-2\right)^2=1\)
c) \(\left(2x-1\right)^3=-8\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
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27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=1^2\)
\(x-2=1\)
\(x=1+2\)
\(x=3\)
c) \(\left(2x-1\right)^3=\left(-8\right)\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=\left(-2\right)\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}\)
27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^2=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)
16 tháng 8 2019
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
16 tháng 8 2019
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
26 tháng 6 2018
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
JK
10 tháng 10 2019
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
10 tháng 10 2019
a) \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)
⇔ \(6x^2-13x+5-6x^2-11x+2=0\)
⇔ \(24x=7\)⇔\(x=\frac{7}{24}\)
b) \(\left(3x-2\right)\left(3x+2\right)-\left(3x-1\right)^2=-5\)
⇔ \(9x^2-4-9x^2+6x-1=5\)
⇔ \(6x=10\)⇔ \(x=\frac{5}{3}\)
c) \(x^2=-6x-8\)⇔\(x^2+6x+8=0\)⇔\(\left(x+2\right)\left(x+4\right)=0\)
⇔\(\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=-\frac{1}{4}\)
a>(x-1/2)^2=0
(x-1/2)^2=0^2
=>x-1/2=0
x=1/2
b>(x-2)^2=1
(x-2)^2=1^2
=>x-2=1
x=3
(2x-1)^3=-8
=>(2x-1)^3=-2^3
=>2x-1=-2
2x=-1
x=-0,5
d>(x+1/2)^2=1/16
(x+1/2)^2=(1/4)^2
=>x+1/2=1/4
x=1/4-1/2
x=-1/4
chuc ban may man trong cuoc song!!!!!