1.a
\(A=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{20}+\sqrt{\left(-2\right)^2}\)
\(=\left|2-\sqrt{5}\right|-2\sqrt{5}+2\)
\(=\sqrt{5}-2-2\sqrt{5}+2=-\sqrt{5}\)
b.
\(B=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{3}{x-9}\right):\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\left(\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}+3-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\left(\dfrac{\sqrt{x}+3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+3}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-3}\)
c.
\(3x+y=5\Leftrightarrow y=-3x+5\)
Đồ thị hàm số song song với \(y=-3x+5\) và cắt trục hoành tại \(x=2\) khi:
\(\left\{{}\begin{matrix}a=-3\\b\ne5\\0=2.a+b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-3\\b=6\end{matrix}\right.\)
2.
a. Em tự giải
b.
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=-5\end{matrix}\right.\)
\(T=x_1\left(1+x_1^2\right)+x_2\left(1+x_2^2\right)=x_1+x_2+x_1^3+x_2^3\)
\(=x_1+x_2+\left(x_1+x_2\right)^3-3x_1x_2.\left(x_1+x_2\right)\)
\(=1+\left(1\right)^3-3.\left(-5\right).1=17\)
Câu 2:
a: \(2x^2-x-28=0\)
=>\(2x^2-8x+7x-28=0\)
=>(x-4)(2x+7)=0
=>\(\left[{}\begin{matrix}x-4=0\\2x+7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b: \(x^2-x-5=0\)
=>\(x_1+x_2=-\dfrac{b}{a}=1;x_1x_2=\dfrac{c}{a}=-5\)
\(T=x_1\left(1+x_1^2\right)+x_2\left(1+x_2^2\right)\)
\(=\left(x_1+x_2\right)+\left(x_1^3+x_2^3\right)\)
\(=1+\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=1+1-3\cdot1\cdot\left(-5\right)\)
=2+15
=17
Câu 1:
a: \(A=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{20}+\sqrt{\left(-2\right)^2}\)
\(=\left|2-\sqrt{5}\right|-2\sqrt{5}+2\)
\(=\sqrt{5}-2-2\sqrt{5}+2=-\sqrt{5}\)
b: \(B=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{3}{x-9}\right):\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+3-3}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}-3}\)
c: 3x+y=5
=>y=-3x+5
Vì đồ thị hàm số y=ax+b song song với đường thẳng y=-3x+5 nên \(\left\{{}\begin{matrix}a=-3\\b\ne5\end{matrix}\right.\)
vậy: y=-3x+b
Thay x=2 và y=0 vào y=-3x+b, ta được:
\(b-3\cdot2=0\)
=>b-6=0
=>b=6
vậy: y=-3x+6
