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Theo vi ét: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=8\end{matrix}\right.\)
Theo đề:
\(B=\dfrac{x_1\sqrt{x_1}-x_2\sqrt{x_2}}{x_1-x_2}=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(x_1+\sqrt{x_1x_2}+x_2\right)}{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(\sqrt{x_1}+\sqrt{x_2}\right)}\left(x_1,x_2\ge0\right)\)
\(=\dfrac{6+\sqrt{8}}{\sqrt{x_1}+\sqrt{x_2}}\)
Tính: \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=x_1+x_2+2\sqrt{x_1x_2}=6+2\sqrt{8}=6+4\sqrt{2}=\left(\sqrt{4}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x_1}+\sqrt{x_2}=\sqrt{4}+\sqrt{2}\) (thỏa mãn \(x_1,x_2\ge0\))
Khi đó: \(P=\dfrac{6+\sqrt{8}}{\sqrt{4}+\sqrt{2}}=4-\sqrt{2}\)
a: x1+x2=-2; x1x2=-4
x1+x2+2+2=-2+2+2=2
(x1+2)(x2+2)=x1x2+2(x1+x2)+4
=-4+2*(-2)+4=-4
Phương trình cần tìm là x^2-2x-4=0
b: \(\dfrac{1}{x_1+1}+\dfrac{1}{x_2+1}=\dfrac{x_1+x_2+2}{\left(x_1+1\right)\left(x_2+1\right)}\)
\(=\dfrac{x_1+x_2+2}{x_1x_2+\left(x_1+x_2\right)+1}\)
\(=\dfrac{-2+2}{-4+\left(-2\right)+1}=0\)
\(\dfrac{1}{x_1+1}\cdot\dfrac{1}{x_2+1}=\dfrac{1}{x_1x_2+x_1+x_2+1}=\dfrac{1}{-4-2+1}=\dfrac{-1}{5}\)
Phương trình cần tìm sẽ là; x^2-1/5=0
c: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{x_1^2+x_2^2}{x_1x_2}=\dfrac{\left(-2\right)^2-2\cdot\left(-4\right)}{-4}=\dfrac{4+8}{-4}=-3\)
x1/x2*x2/x1=1
Phương trình cần tìm sẽ là:
x^2+3x+1=0
\(x^2-4x-6=0\)
\(\text{Δ}=\left(-4\right)^2-4\cdot1\cdot\left(-6\right)=16+24=40>0\)
=>Phương trình này có hai nghiệm phân biệt
Theo vi-et, ta có:
\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-6}{1}=-6\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=4^2-2\cdot\left(-6\right)=16+12=28\)
\(B=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1\cdot x_2}=\dfrac{4}{-6}=-\dfrac{2}{3}\)
\(C=x_1^3+x_2^3\)
\(=\left(x_1+x_2\right)^3-3\cdot x_1\cdot x_2\cdot\left(x_1+x_2\right)\)
\(=4^3-3\cdot4\cdot\left(-6\right)=64+72=136\)
\(D=\left|x_1-x_2\right|\)
\(=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{4^2-4\cdot\left(-6\right)}=\sqrt{16+24}=\sqrt{40}=2\sqrt{10}\)
Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{1}{1}=1\\x_1x_2=\dfrac{c}{a}=-\dfrac{3}{1}=-3\end{matrix}\right.\)
a
\(A=x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2=1^2-2.\left(-3\right)=1+6=7\)
b
\(B=x_1^2x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)=\left(-3\right).1=-3\)
c
\(C=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
d
\(D=\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=\dfrac{x_2^2}{x_1x_2}+\dfrac{x_1^2}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{1^2-2.\left(-3\right)}{-3}=\dfrac{1+6}{-3}=\dfrac{7}{-3}=-\dfrac{3}{7}\)
(a) Khi \(m=2,\left(1\right)\Leftrightarrow x^2-4x-5=0\left(2\right)\).
Phương trình (2) có \(a-b+c=1-\left(-4\right)+\left(-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{c}{a}=5\end{matrix}\right.\).
Vậy: Khi \(m=2,S=\left\{-1;5\right\}\).
(b) Điều kiện: \(x_1,x_2\ne0\Rightarrow m\in R\)
Phương trình có nghiệm khi:
\(\Delta'=\left(-m\right)^2-1\cdot\left(-m^2-1\right)\ge0\)
\(\Leftrightarrow2m^2+1\ge0\left(LĐ\right)\)
Suy ra, phương trình (1) có nghiệm với mọi \(m\).
Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=-m^2-1\end{matrix}\right.\)
Theo đề: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-\dfrac{5}{2}\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=-\dfrac{5}{2}\)
\(\Leftrightarrow2\left(x_1+x_2\right)^2+x_1x_2=0\)
\(\Leftrightarrow2\left(2m\right)^2+\left(-m^2-1\right)=0\)
\(\Leftrightarrow7m^2=1\Leftrightarrow m=\pm\dfrac{\sqrt{7}}{7}\) (thỏa mãn).
Vậy: \(m=\pm\dfrac{\sqrt{7}}{7}.\)
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-5}{3}\\x_1x_2=\dfrac{2}{3}\end{matrix}\right.\)
\(Q=\dfrac{x_1}{x_2+2}+\dfrac{x_2}{x_1+2}\)
\(\Rightarrow Q=\dfrac{x_1\left(x_1+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}+\dfrac{x_2\left(x_2+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}\)
\(\Rightarrow Q=\dfrac{x^2_1+2x_1+x^2_2+2x_2}{x_1x_2+2x_1+2x_2+4}\)
\(\Rightarrow Q=\dfrac{\left(x^2_1+x^2_2\right)+\left(2x_1+2x_2\right)}{x_1x_2+\left(2x_1+2x_2\right)+4}\)
\(\Rightarrow Q=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
\(\Rightarrow Q=\dfrac{\left(-\dfrac{5}{3}\right)^2-2.\dfrac{2}{3}+2\left(\dfrac{-5}{3}\right)}{\dfrac{2}{3}+2\left(\dfrac{-5}{3}\right)+4}\)
\(\Rightarrow Q=\dfrac{-17}{12}\)
\(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(m^2-3\right)\)
\(=4m^2+8m+4-4m^2+12=8m+16\)
Để phương trình có hai nghiệm thì 8m+16>=0
hay m>=-2
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2-3\end{matrix}\right.\)
Theo đề, ta có: \(x_1^2+x_2^2+1=3x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2+1=0\)
\(\Leftrightarrow\left(2m+2\right)^2-5\left(m^2-3\right)+1=0\)
\(\Leftrightarrow4m^2+8m+4-5m^2+15+1=0\)
\(\Leftrightarrow-m^2+8m+20=0\)
=>(m-10)(m+2)=0
=>m=10 hoặc m=-2
a, \(\Delta'=\left(m+1\right)^2-\left(m^2-3\right)=m^2+2m+1-m^2+3=2m+4\)
Để pt có 2 nghiệm x1 ; x2 khi \(\Delta'\ge0\Leftrightarrow m\ge-2\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2-3\end{matrix}\right.\)
Ta có : \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}+\dfrac{1}{x_1x_2}=3\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+1}{x_1x_2}=3\)
\(\Leftrightarrow\dfrac{4\left(m^2+2m+1\right)-2\left(m^2-3\right)+1}{m^2-3}=3\)
\(\Rightarrow2m^2+8m+11=3m^2-9\Leftrightarrow m^2-8m-20=0\Leftrightarrow m=10;m=-2\)(tm)
Ptrình : \(x^2-7x+10=0\)
Ta có : \(\Delta=\left(-7\right)^2-4.1.10=9>0\)
=> Phương trình có 2 nghiệm phân biệt \(x1\) và \(x2\)
\(x1=\dfrac{-\left(-7\right)+\sqrt{\Delta}}{2.1}=\dfrac{7+\sqrt{9}}{2}=5\)
\(x2=\dfrac{-\left(-7\right)-\sqrt{\Delta}}{2.1}=\dfrac{7-\sqrt{9}}{2}=2\)
Vậy :
A = \(x_1^2+x_2^2+3x_1x_2=5^2+2^2+3.5.2=59\)
B = .................
.... (có x1 và x2 rồi thik thay vào lak tính đc, cái này bn tự tính nha)