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a: A=x1+x2=-5/2
b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)
c: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)
\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)
e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-1\\x_1x_2=-2+\sqrt{2}\end{matrix}\right.\)
\(A=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-1}{-2+\sqrt{2}}=\dfrac{2+\sqrt{2}}{2}\)
\(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(-1\right)^2-2\left(-2+\sqrt{2}\right)=5-2\sqrt{2}\)
2:
\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)
B=(x1+x2)^2-2x1x2
=3^2-2*(-7)
=9+14=23
C=căn (x1+x2)^2-4x1x2
=căn 3^2-4*(-7)=căn 9+28=căn 27
D=(x1^2+x2^2)^2-2(x1x2)^2
=23^2-2*(-7)^2
=23^2-2*49=431
D=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=69+10*(-7)=-1
\(x^2-4x-6=0\)
\(\text{Δ}=\left(-4\right)^2-4\cdot1\cdot\left(-6\right)=16+24=40>0\)
=>Phương trình này có hai nghiệm phân biệt
Theo vi-et, ta có:
\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-6}{1}=-6\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=4^2-2\cdot\left(-6\right)=16+12=28\)
\(B=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1\cdot x_2}=\dfrac{4}{-6}=-\dfrac{2}{3}\)
\(C=x_1^3+x_2^3\)
\(=\left(x_1+x_2\right)^3-3\cdot x_1\cdot x_2\cdot\left(x_1+x_2\right)\)
\(=4^3-3\cdot4\cdot\left(-6\right)=64+72=136\)
\(D=\left|x_1-x_2\right|\)
\(=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{4^2-4\cdot\left(-6\right)}=\sqrt{16+24}=\sqrt{40}=2\sqrt{10}\)
Câu 1
a) Xét phương trình : 2x2 +5x - 8 = 0
Có \(\Delta=5^2-4.2.\left(-8\right)=89>0\)
=> Phương trình luôn có 2 nghiệm phân biệt x1, x2
b) Do phương trình luôn có 2 nghiệm x1,x2
=> Theo định lí viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{2}\\x_1.x_2=-4\end{matrix}\right.\)
A = \(\dfrac{2}{x_1}+\dfrac{2}{x_2}=\dfrac{2.x_2}{x_1x_2}+\dfrac{2x_1}{x_1x_2}=\dfrac{2\left(x_1+x_2\right)}{x_1x_2}=\dfrac{2.\left(-\dfrac{5}{2}\right)}{-4}=\dfrac{-5}{-4}=\dfrac{5}{4}\)
Vậy A = \(\dfrac{5}{4}\)
Câu 2
Ta có \(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{x}+2}+\dfrac{4-a}{2-\sqrt{a}}\left(a\ge0;a\ne4\right)\)
\(=\dfrac{\left(2+\sqrt{a}\right)^2}{2+\sqrt{a}}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\sqrt{a}+2+\left(2+\sqrt{a}\right)=2\sqrt{a}+4\)
Vậy P = \(2\sqrt{a}+4\left(a\ge0;a\ne4\right)\)
b) Ta có a2 - 7a + 12 = 0
\(\Leftrightarrow a^2-4a-3a+12=0\)
\(\Leftrightarrow a\left(a-4\right)-3\left(a-4\right)=0\)
\(\Leftrightarrow\left(a-4\right)\left(a-3\right)=0\Leftrightarrow\left[{}\begin{matrix}a=4\left(loại\right)\\a=3\end{matrix}\right.\)
Với a = 3 thay vào P ta được P = \(2\sqrt{3}+4\)
\(\Rightarrow\sqrt{P}=\sqrt{2\sqrt{3}+4}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
Vậy \(\sqrt{P}=\sqrt{3}+1\) tại a2 -7a + 12 =0
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)Có ac=-1<0
=>pt luôn có hai nghiệm trái dấu
b)Do x1;x2 là hai nghiệm của pt
=> \(\left\{{}\begin{matrix}x_1^2-mx_1-1=0\\x_2^2-mx_2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-1=mx_1\\x_2^2-1=mx_2\end{matrix}\right.\)
=>\(P=\dfrac{mx_1+x_1}{x_1}-\dfrac{mx_2+x_2}{x_2}\)\(=m+1-\left(m+1\right)=0\)
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
