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\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
Vậy \(M=x^2+11xy-y^2\)
\(b.M+(3x^2y-2xy^3)=2x^2y-4xy^3\)
\(M=(2x^2y-4xy^3)-(3x^2-2xy^3)\)
\(M=
\) \(2x^2-4xy^3-3x^2+2xy^3\)
\(M=(2x^2-3x^2)+(-4xy^3+2xy^3)\)
\(M=-x^2-2xy^3\)
Vậy \(M=-x^2-2xy^3\)
a) M + (5x\(^2\) - 2xy) = 6x\(^2\) + 9xy - y\(^2\)
=> M = (6x\(^2\) + 9xy - y\(^2\)) - (5x\(^2\) - 2xy)
M = 6x\(^2\) + 9xy - y\(^2\) - 5x\(^2\) + 2xy
M = (6x\(^2\) - 5x\(^2\)) + (9xy + 2xy) - y\(^2\)
M = 1x\(^2\) + 11xy - y\(^2\)
a) Các đơn thức đồng dạng là:
\(2x^2y^3\) và \(\dfrac{-1}{2}x^2y^3\); \(5y^2x^3\) và \(\dfrac{-1}{2}x^2y^3\)
b) Ta có:
\(F=2x^2y^3+5y^2x^3+\dfrac{-1}{2}x^3y^2+\dfrac{-1}{2}x^2y^3\)
\(=\left(2x^2y^3+\dfrac{-1}{2}x^2y^3\right)+\left(5y^2x^3+\dfrac{-1}{2}x^3y^2\right)\)
\(=\dfrac{3}{2}x^2y^3+\dfrac{9}{2}x^3y^2\)
c) Tại \(x=-3;y=2\) thì:
\(F=\dfrac{3}{2}\left(-3\right)^2.2^3+\dfrac{9}{2}.\left(-3\right)^3.2^2\)
\(=108-486=-378.\)
a) Các đơn thức đồng dạng là:
- 2x\(^2\)y\(^3\) và \(\dfrac{-1}{2}\)x\(^2\)y\(^3\)
- 5y\(^2\)x\(^3\) và \(\dfrac{-1}{2}\)x\(^3\)y\(^2\)
b) F = 2x\(^2\)y\(^3\) + 5y\(^2\)x\(^3\) + (\(\dfrac{-1}{2}\))x\(^3\)y\(^2\) + (\(\dfrac{-1}{2}\))x\(^2\)y\(^3\)
= [ 2x\(^2\)y\(^3\) + (\(\dfrac{-1}{2}\))x\(^2\)y\(^3\) ] + [ 5y\(^2\)x\(^3\) + (\(\dfrac{-1}{2}\))x\(^3\)y\(^2\) ]
= \(\dfrac{3}{2}\)x\(^2\)y\(^3\) + \(\dfrac{9}{2}\)y\(^2\)x\(^3\)
Vậy đa thức F có giá trị là: \(\dfrac{3}{2}\)x\(^2\)y\(^3\) + \(\dfrac{9}{2}\)y\(^2\)x\(^3\)
c) Thay x = -3 và y = 2 vào đa thức F đã cho, ta có:
\(\dfrac{3}{2}\) . (-3)\(^2\) . 2\(^3\) + \(\dfrac{9}{2}\) . 2\(^2\) . (-3)\(^3\) = 108 + (-486) = 108 - 486 = -378
Vậy tại x = -3 và y = 2, giá trị của đa thức F là: -378
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
=> \(7-3x=0\) hoặc \(2x+1=0\)
\(3x=7-0\) hoặc \(2x=0-1\)
\(3x=7\) hoặc \(2x=-1\)
\(x=7:3\) hoặc \(x=-1:2\)
\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)
Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
Kêu người ta giúp mà ói vào mặt người ta vậy à?
E + F = (5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1) + (2x\(^2\)y - xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\))
= 5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1 + 2x\(^2\)y -xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\)
= (5xy - \(\dfrac{2}{5}\)xy) + (\(\dfrac{-2}{3}\)x\(^2\)y + 2x\(^2\)y) + (xyz\(^2\) - xyz\(^2\)) + (-1 + \(\dfrac{1}{2}\)) + x
= \(\dfrac{23}{5}\)xy + \(\dfrac{4}{3}\) x\(^2\)y - \(\dfrac{1}{2}\) + x
E - F = (5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1) - (2x\(^2\)y - xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\))
= 5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1 - 2x\(^2\)y + xyz\(^2\) + \(\dfrac{2}{5}\)xy - x - \(\dfrac{1}{2}\)
= (5xy + \(\dfrac{2}{5}\)xy) + (\(\dfrac{-2}{3}\)x\(^2\)y - 2x\(^2\)y) + (xyz\(^2\) + xyz\(^2\))+ (-1 - \(\dfrac{1}{2}\)) - x
= \(\dfrac{27}{5}\)xy - \(\dfrac{8}{3}\)x\(^2\)y + 2xyz\(^2\) - \(\dfrac{3}{2}\) - x
Vậy E - F = \(\dfrac{27}{5}\)xy - \(\dfrac{8}{3}\)x\(^2\)y + 2xyz\(^2\) - \(\dfrac{3}{2}\) - x