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a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
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)
\(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)
Bài 1:
a,\(0,75+\frac{9}{17}-1\frac{4}{5}-\frac{26}{17}-2\frac{4}{5}\)
\(=\frac{3}{4}+\left(\frac{9}{17}-\frac{26}{17}\right)-\left(1\frac{4}{5}+2\frac{4}{5}\right)\)
\(=\frac{3}{4}-1-\frac{23}{5}\)
\(=\frac{15}{20}-\frac{20}{20}-\frac{92}{20}=\frac{-97}{20}\)
Bài 2:
a, \(\left(2x+\frac{3}{4}\right)-\frac{10}{3}=\frac{-13}{3}\)
\(2x+\frac{3}{4}=\frac{-13}{3}+\frac{10}{3}\)
\(2x+\frac{3}{4}=-1\)
\(2x=-1-\frac{3}{4}\)
\(2x=\frac{-7}{4}\)
x = -7/8
b, 3,2x - 2,7x + 8,5 = 6
x(3,2 - 2,7) = -2,5
0,5x = -2,5
x = -5
a) \(\left|x-3\right|+\left|2x-6\right|=8\)
\(x-3+2x-6=8\)
\(3x-9=8\)
\(3x=17\)
\(\Rightarrow x=\frac{17}{3}\)
b) Tương tự câu a .
c) \(\left|2x-3\right|=6-\left|3-2x\right|\)
\(2x-3=6-3-2x\)
\(2x-3=x\)
\(-2x=3\)
\(x=\frac{-3}{2}\)
d) \(\left|3x-2\right|-\left|6-9x\right|=-\left|-16\right|\)
\(3x-2-6-9x=-16\)
\(3x-8-9x=-16\)
\(-6x-8=-16\)
\(-6x=-8\)
\(\Rightarrow x=\frac{8}{6}\)
\(\)
a, x=17/8 hoặc x=7/8
b, x= +- 2,7
c, x=7/30 hoặc
x= -43/30
d, x= 1/2 hoặc x= -7/2
Chức bạn học tốt!!
thankyou bn nha