a) |x|+0.573=2
b) |x+1/3|-4=-1
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KL
1
N
1
20 tháng 1 2022
a: \(\Leftrightarrow\left|x\cdot\dfrac{1}{3}-\dfrac{1}{4}\right|=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{2}{3}\\x\cdot\dfrac{1}{3}-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{4}{7}+\dfrac{3}{4}\right|=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{4}{7}+\dfrac{3}{4}=\dfrac{1}{6}\\x\cdot\dfrac{4}{7}+\dfrac{3}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-49}{48}\\x=-\dfrac{77}{48}\end{matrix}\right.\)
T
13 tháng 8 2023
a) (3x - 2)2 - (1 + 5x)2
= (3x - 2 - 1 - 5x)(3x - 2 + 1 + 5x)
= (-2x - 3)(8x - 1)
b) (3x + 4)(3x - 4) - (5 - x)2
= (3x)2 - 42 - (25 - 10x + x2)
= 9x2 - 16 - 25 + 10x - x2
= 8x2 + 10x - 41
c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)
\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)
\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)
\(=4x+25\)
13 tháng 8 2023
a: =9x^2-12x+4-25x^2-10x-1
=-16x^2-22x+3
b: =9x^2-16-x^2+10x-25
=8x^2+10x-41
c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)
KN
4
WT
0
12 tháng 8 2021
a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)
\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)
\(\Leftrightarrow3-x-2+7x-7=0\)
\(\Leftrightarrow6x-6=0\)
hay x=1(loại
b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)
\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)
Suy ra: \(-2-x^2+5x-4=x^2+x-6\)
\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
12 tháng 8 2021
\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)
\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow3-x-2+7x-7=0\)
\(\Rightarrow6x-6=0\)
\(\Rightarrow x=1\)
a, <=> |x| = 1,427
<=> x=1,427 hoặc x= -1,427
b, <=> |x+1/3| = 5
<=> x+1/3= 5 hoặc x+1/3 = -5
<=> x= 14/3 hoặc x= -16/3
a) đúng
b) sai