cần gấp ạa
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KH
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
P
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) \(Q=\dfrac{x_1}{x_2+2}+\dfrac{x_2}{x_1+2}\)
\(Q=\dfrac{x_1\left(x_1+2\right)+x_2\left(x_2+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}\)
\(Q=\dfrac{x^2_1+2x_1+x_2^2+2x_2}{x_1x_2+2x_2+2x_1+4}\)
\(Q=\dfrac{\left(x^2_1+x^2_2\right)+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(Q=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(Q=\dfrac{\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}+2\cdot-\dfrac{5}{3}}{\dfrac{2}{3}+2\cdot-\dfrac{5}{3}+4}=-\dfrac{17}{12}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
10 tháng 12 2021
Bài 33:
a: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 11 2021
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
NA
0
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)