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1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
1) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x.1+5^x.5^2=650\)
\(\Rightarrow5^x.\left(1+5^2\right)=650\)
\(\Rightarrow5^x.26=650\)
\(\Rightarrow5^x=650:26\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Mình chỉ làm câu 1) thôi nhé.
Chúc bạn học tốt!
a) (-xy)10 : (-xy)5 = (-xy)2
b) (5x2y4) : 10x2y = y3/2
c) (15x4y3z2) : (5x2y2z2) = 3x2y
d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}x^2y^2\right)=\frac{\frac{3}{4}x^3y^2}{-\frac{1}{2}x^2y^2}=-\frac{3}{2}x\)
e) 6x3y5 : 12x3y2 = y3/2
f) (25x5 - 5x4 + 10x2) : (5x2)
= 5x3 - x2 + 2
g) \(\frac{2}{3}xy\cdot\left(2x^2y-3xy+y^2\right)=\frac{4}{3}x^3y^2-2x^2y^2+\frac{2}{3}xy^3\)
h) \(\frac{3}{4}x^3y^5z:\left(5x^2y^2z\right)=\frac{\frac{3}{4}x^3y^5z}{5x^2y^2z}=\frac{3}{20}xy^3\)
a) \(\frac{2x}{3}=\frac{3y}{4}\Leftrightarrow8x=9y\Rightarrow x=\frac{9y}{8}\left(1\right)\)
\(\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow15y=16z\Rightarrow z=\frac{15y}{16}\left(2\right)\)
THay (1) và (2) vào biểu thức \(x+y+z=41\);ta được : \(\frac{9y}{8}+y+\frac{15y}{16}=41\)
\(\Rightarrow18y+16y+15y=656\Rightarrow y=\frac{656}{49}\)
Do đó : \(x=\frac{\frac{9.656}{49}}{8}=\frac{738}{49}\)
\(z=\frac{\frac{15.656}{49}}{16}=\frac{615}{49}\)
KL : \(x=\frac{738}{49};y=\frac{656}{49};z=\frac{615}{49}\)
b) Ta có : \(4x=3y\Rightarrow x=\frac{3y}{4}\)(1)
\(5y=6z\Rightarrow z=\frac{5y}{6}\)(2)
Thay (1) và (2) vào biểu thức \(x^2+y^2+z^2=500\);ta được :
\(\left(\frac{3y}{4}\right)^2+y^2+\left(\frac{5y}{6}\right)^2=500\)
\(\Rightarrow\frac{9y^2}{16}+y^2+\frac{25y^2}{36}=500\Rightarrow324y^2+576y^2+400y^2=288000\)
\(\Rightarrow1300y^2=288000\Rightarrow y^2=\frac{2880}{13}\Rightarrow\orbr{\begin{cases}y=\frac{24\sqrt{65}}{13}\\y=-\frac{24\sqrt{65}}{13}\end{cases}}\)
Với \(y=\frac{24\sqrt{65}}{13}\Rightarrow x=\frac{3\cdot\frac{24\sqrt{65}}{13}}{4}=\frac{18\sqrt{65}}{13};z=\frac{5\cdot\frac{24\sqrt{65}}{13}}{6}\)
\(y=-\frac{24\sqrt{65}}{13}\Rightarrow x=-\frac{18\sqrt{65}}{13};z=\frac{5\cdot-\frac{24\sqrt{65}}{13}}{6}\)
TRA LOI
BẠN CHỈ CẦN NHỚ 7 ĐẲNG THÚC
HỌC TÔT
\(x^2-4xy-y^2=\left(x^2-4xy+y^2\right)-2y^2\)
\(=\left(x-y\right)^2-\left(\sqrt{2}.y\right)^2\)
\(=\left(x-y-\sqrt{2}.y\right)\left(x-y+\sqrt{2}y\right)\)
\(x^4-\frac{1}{2}=\left(x^2\right)^2-\left(\sqrt{\frac{1}{2}}\right)^2=\left(x^2-\sqrt{\frac{1}{2}}\right)\left(x^2+\sqrt{\frac{1}{2}}\right)\)