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a: =>(3/2-2x):2/3=1/6
=>3/2-2x=1/6x2/3=2/18=1/9
=>2x=25/18
hay x=25/36
b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)
=>x-1/4=1/2
=>x=3/4
c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)
=>23/12x=2/3
=>x=8/23
`#040911`
a,
\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)
\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)
Vậy, \(x=-\dfrac{8}{21}\)
b,
\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, \(x\in\left\{-2;3\right\}\)
c,
\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)
Bạn xem lại đề có sai kh nhỉ?
c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)
\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)
\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
\(a,\frac{1}{3}+\frac{1}{2}:x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{1}{5}-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{15}-\frac{5}{15}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{-2}{15}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{-15}{2}=\frac{-15}{4}\)
\(b,\frac{1}{3}x+\frac{2}{5}\left[x+1\right]=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=\frac{-2}{5}:\frac{11}{15}=\frac{-2}{5}\cdot\frac{15}{11}=\frac{-2}{1}\cdot\frac{3}{11}=\frac{-6}{11}\)
a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);
b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);
c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) = - 3{x^2}.6{x^2} - - 3{x^2}.8x + - 3{x^2}.1\\ = - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} = - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);
d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);
e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ = - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} = - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);
g)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21