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1a) \(10^{n+1}-6\cdot10^n\)
\(=10^n\cdot10-6\cdot10^n\)
= \(10^n\left(10-6\right)\)
\(=10^n\cdot4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n\cdot2^3+2^n\cdot2^2-2^n\cdot2+2^n\)
\(=2^n\left(2^3+2^2-2+1\right)\)
\(=2^n\cdot11\)
c) \(90\cdot10^k-10^{k+2}+10^{k+1}\)
\(=90\cdot10^k-10^k\cdot10^2+10^k\cdot10\)
\(=10^k\left(90-10^2+10\right)=0\)
d) \(2,5\cdot5^{n-3}\cdot10+5^n-6\cdot5^{n-1}\)
\(=\dfrac{2,5\cdot10\cdot5^n}{5^3}+5^n-\dfrac{6\cdot5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6\cdot5^n}{5}\)
\(=\dfrac{5^n+5^n\cdot5-6\cdot5^n}{5}=\dfrac{5^n\left(5-6\right)+5^n}{5}=0\)
2. \(M+\left(6x^2-4xy\right)=7x^2-8xy+y^2\)
\(M=\left(7x^2-8xy+y^2\right)-\left(6x^2-4xy\right)\)
\(M=7x^2-8xy+y^2-6x^2+4xy\)
\(M=7x^2-6x^2-8xy+4xy+y^2\)
\(M=x^2-4xy+y^2\)
Bài 1:
a) \(x^2+7x-8=x^2+2.x.\frac{7}{2}+\frac{49}{4}-\frac{81}{4}\)
\(=\left(x+\frac{7}{2}\right)^2-\frac{81}{4}=0\)
\(\Rightarrow\left(x+\frac{7}{2}\right)^2=\frac{81}{4}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{7}{2}=\frac{9}{2}\\x+\frac{7}{2}=\frac{-9}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-8\end{cases}}\)
Vậy nghiệm của đa thức m(x) là 1 hoặc -8
b) \(\left(x-3\right)\left(16-4x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\16-4x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
Vậy nghiệm của đa thức g(x) là 3 hoặc 4
c) \(5x^2+9x+4=0\)
\(\Rightarrow x^2+\frac{9}{5}x+\frac{4}{5}=0\)
\(\Rightarrow x^2+2x.\frac{9}{10}+\frac{81}{100}-\frac{1}{100}=0\)
\(\Rightarrow\left(x+\frac{9}{10}\right)^2-\frac{1}{100}=0\)
\(\Rightarrow\left(x+\frac{9}{10}\right)^2=\frac{1}{100}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{9}{10}=\frac{1}{10}\\x+\frac{9}{10}=\frac{-1}{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=-1\end{cases}}\)
Vậy...
ta có: -1 là nghiệm của M(y)
=> a.(-1)^2 + b. (-1) +c = 0
a - b +c =0 => a - ( b-c) =0 => a = b-c
mà 5a + b +2c =0
=> a - b+ c = 5a +b+2c = 0
( b-c) + b +c = 5.( b-c) + b+ 2c =0
b-c + b+ c = 5b - 5c + b + 2c =0
=> 2c = 6b - 3c =0
=> 2c =3. ( 2b - c) =0
=> 2c =0 => c =0
=> 3 .( 2b - c) = 0 => 2b -c =0 => 2b - 0 =0 => 2b =0 => b =0
mà a = b-c
=> a = 0-0
=> a =0
KL: a =b=c =0
mk nghĩ như z đó!!!
a, P + 3x\(^{^2}\) - 4xy = 6y\(^{^2}\) - 9xy + x\(^2\)
=> P = 6y\(^2\)- 9xy + x\(^2\)+ 4xy - 3x\(^2\)= 6y\(^2\)- 5xy - 2x\(^2\)
=> P = 6y\(^2\) - 5xy - 2x\(^2\)
b,
4y\(^2\) - 8xy - P = 5x\(^2\) - 12xy + 4y\(^2\)
=> P = 4y\(^2\) - 8xy - 5x\(^2\) + 12xy - 4y\(^2\) = 4xy - 5x\(^2\)
=> P = 4xy - 5x\(^2\)
c,
P - ( x\(^2\) - 2y\(^2\) + 3z\(^2\) ) + 3x\(^2\) - y\(^2\) + 2z\(^2\)= 2x\(^2\) - 3y\(^2\) -z\(^2\)
= P + 2x\(^2\) + y\(^2\) - z\(^2\) = 2x\(^2\) - 3y\(^2\) - z\(^2\)
=> P = 2x\(^2\) - 3y\(^2\) - z\(^2\) - 2x\(^2\) - y\(^2\) + z\(^2\)
=> P = -2y\(^2\)
\(a.M+(5x^2-2xy)=6x^2+9xy-y^2
\)
\(M=(6x^2+9xy-y^2)-(5x^2-2xy)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=(6x^2-5x^2)+(9xy+2xy)-y^2\)
\(M=x^2+11xy-y^2\)
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\(^2\)
\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\\ \Rightarrow2f\left(x\right)=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\\ 2f\left(x\right)=10x^4-6x^3+4x^2+8x-14\\ 2f\left(x\right)=2\left(5x^4-3x^3+2x^2+4x-7\right)\\ \Rightarrow f\left(x\right)=5x^4-3x^3+2x^2+8x-14\)
\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\\ \Rightarrow g\left(x\right)=6x^4-3x^2-5-f\left(x\right)\\ g\left(x\right)=6x^4-3x^2-5-5x^4+3x^3-2x^2-8x+14\\ g\left(x\right)=x^4+3x^3-5x^2-8x+9\)
Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\left(\forall x\right)\\\left(3y+4\right)^{2020}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\left(\forall x,y\right)\)
Mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\left(\forall x,y\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Khi đó thay vào ta được:
\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)
\(\Rightarrow M=-\frac{1159}{36}\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=x^2+11xy-y^2\)
\(N=3xy-4y^2-x^2+7xy-8y^2\)
\(N=-x^2+10xy-12y^2\)
a. \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(\Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\)
b. \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)
\(\Rightarrow N=3xy-4y^2-x^2+7xy-8y^2\)
a, \(M=7.\left(x-y\right)+4a.\left(x-y\right)-5\)
Theo bài ra ta có: x-y=0
=> \(M=0+0-5\)
\(\Rightarrow M=-5\)
b,
\(N=\left(x^2+y^2\right).\left(x-y\right)+3\)
\(\Rightarrow N=0+3=3\)
lớp 7 lên 8 à làm quen nhá :)
a) \(M=7x-7y+4ax-4ay-5\)
\(M=7\left(x-y\right)+4a\left(x-y\right)-5\)
\(M=0+0-5=-5\)
b) \(N=x\left(x^2+y^2\right)-y\left(x^2+y^2\right)+3\)
\(N=\left(x-y\right)\left(x^2+y^2\right)+3\)
\(N=0+3=3\)
\(M+\left(6x^2-4xy\right)=7x^2-8xy+y^2\\ \Rightarrow M=7x^2-8xy+y^2-\left(6x^2-4xy\right)\\ =7x^2-8xy+y^2-6x^2+4xy\\ =\left(7x^2-6x^2\right)+\left(-8xy+4xy\right)+y^2\\ =x^2-4xy+y^2\)