cần gấp ạa
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mik cần gấp ạa
1
KH
1
25 tháng 10 2021
16, Jane hasn't cooked for her family since she was at high school
17, The last time Jane cooked for her family was when she was at high school
12 tháng 1 2024
Ta có: \(-3x^2-5x-2=0\)
Theo định lý vi-et ta có:
\(x_1+x_2=-\dfrac{b}{a}=-\dfrac{-5}{-3}=-\dfrac{5}{3}\)
\(x_1x_2=\dfrac{c}{a}=\dfrac{-2}{-3}=\dfrac{2}{3}\)
a) \(M=x_1+\dfrac{1}{x_1}+\dfrac{1}{x_2}+x_2\)
\(M=\left(x_1+x_2\right)+\dfrac{x_1+x_2}{x_1x_2}\)
\(M=-\dfrac{5}{3}+\dfrac{-\dfrac{5}{3}}{\dfrac{2}{3}}=-\dfrac{25}{6}\)
b) \(N=\dfrac{1}{x_1+3}+\dfrac{1}{x_2+3}\)
\(N=\dfrac{x_2+3+x_1+3}{\left(x_1+3\right)\left(x_2+3\right)}\)
\(N=\dfrac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}\)
\(N=\dfrac{-\dfrac{5}{3}+6}{\dfrac{2}{3}+3\cdot-\dfrac{5}{3}+9}=\dfrac{13}{14}\)
c) \(P=\dfrac{x_1-3}{x^2_1}+\dfrac{x_2-3}{x^2_2}\)
\(P=\dfrac{x^2_2\left(x_1-3\right)+x^2_1\left(x_2-3\right)}{x^2_1x^2_2}\)
\(P=\dfrac{x^2_2x_1+x^2_1x_2-3x^2_2-3x^2_1}{\left(x_1x_2\right)^2}\)
\(P=\dfrac{x_1x_2\left(x_1+x_2\right)-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{\left(x_1x_2\right)^2}\)
\(P=\dfrac{\dfrac{2}{3}\cdot-\dfrac{5}{3}-3\cdot\left[\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}\right]}{\left(\dfrac{2}{3}\right)^2}=-\dfrac{49}{4}\)
d) \(Q=\dfrac{x_1}{x_2+2}+\dfrac{x_2}{x_1+2}\)
\(Q=\dfrac{x_1\left(x_1+2\right)+x_2\left(x_2+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}\)
\(Q=\dfrac{x^2_1+2x_1+x_2^2+2x_2}{x_1x_2+2x_2+2x_1+4}\)
\(Q=\dfrac{\left(x^2_1+x^2_2\right)+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(Q=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(Q=\dfrac{\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}+2\cdot-\dfrac{5}{3}}{\dfrac{2}{3}+2\cdot-\dfrac{5}{3}+4}=-\dfrac{17}{12}\)
10 tháng 12 2021
Bài 33:
a: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)
12 tháng 11 2021
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
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MD
Mik cần gấp lms luôn ạa mng giúp mik vớii
0
Bài 3. a) A(x)=2x^2(x-3)-5(x^2-2x-5)
A(x)=2x^3-6x^2-5x^2+10x+25
A(x)=2x^3-11x^2+10x+25
B(x)=x^3-3x(x^2-2x-5)
B(x)=x^3-3x^3+6x^2+15x
B(x)=-2x^3+6x^2+15x
Đa thức A(x) có:
Bậc: 3
HSTD: 25
HSCN:2
b) A(x)-C(x)=B(x)
C(x)=A(x)-B(x)=2x^3-11x^2+10x+25+2x^3-6x^2-15x
C(x)=4x^3-17x^2-5x+25
c) P(x)=B(x)+2x^3=-2x^3+6x^2+15x+2x^3=6x^2+15x
Ta có: 6x^2+15x=0
x(6x+15)=0
x=0 hoặc 6x+15=0=>x=-5/2
Vậy x=0;x=-5/2 là nghiệm P(x)