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M = ( x^3 + 5xy^2 - y^3 ) - ( x^3 - 2xy^2 + y^3)
M = x^3 + 5xy^2 - y^3 - x^3 + 2xy^2 - y^3
M = ( x^3 - x^3 ) + ( - y^3 - y^3 ) + ( 5xy^2 + 2xy^2)
M = 0 - 2y3 + 7xy2
ai giúp mik câu này vs ak
mik tick cho
M-(xy3-2xy+x2+5)=xy3+5xy-2x2-6
(xy3-2xy+x2+5)-M=xy3+5xy-2x2-6
A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
| x | \(\frac{-2}{3}\) 3 |
| 3x+2 | - 0 + | + |
| x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
a, \(\left(m-2\right)^2=1\)
\(\Rightarrow m-2\in\left\{-1;1\right\}\Rightarrow m\in\left\{1;3\right\}\)
Vậy..............
b, \(\left(2m-1\right)^3=-8\)
\(\Rightarrow2m-1=-2\Rightarrow2m=-1\Rightarrow m=-\dfrac{1}{2}\)
Vậy.................
c, \(\left(m+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow m+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow m\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
Vậy.................
Chúc bạn học tốt!!!
Bài 2:
\(mxy^2-3xy^2+7xy^2=5xy^2\)
=>m-3+7=5
=>m+4=5
hay m=1
\(M=5xy+\dfrac{3}{2}x^2-6y^2+2xy-4y^2+3x^2\)
\(M=\left(5xy+2xy\right)+\left(\dfrac{3}{2}x^2+3x^2\right)+\left(-6y^2-4y^2\right)\)
\(M=7xy+\dfrac{9}{2}y^2-10x^2\)