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a: \(\Leftrightarrow2x^2+6x-3x-9=0\)
=>(x+3)(2x-3)=0
=>x=3/2 hoặc x=-3
b: \(\Leftrightarrow6x\left(1-2x\right)=0\)
=>x=0 hoặc 1-2x=0
=>x=0 hoặc x=1/2
c: \(\Leftrightarrow8x^2=1\)
\(\Leftrightarrow x^2=\dfrac{2}{16}\)
hay \(x\in\left\{\dfrac{\sqrt{2}}{4};-\dfrac{\sqrt{2}}{4}\right\}\)
d: \(\Leftrightarrow x^4-9x^2+2x^2-18=0\)
\(\Leftrightarrow x^2-9=0\)
=>x=3 hoặc x=-3
a: \(P=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
b: P<1
=>P-1<0
=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)
=>căn a-1<0
=>0<a<1
c: Thay x=19-8căn3 vào P, ta được:
\(P=\dfrac{19-8\sqrt{3}+4+\sqrt{3}+1}{4+\sqrt{3}-1}=\dfrac{31-15\sqrt{3}}{2}\)
\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)
\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{8}{1}=8\)
Lm ơn giúp mik đii mà mik bt ơn bn đó nhiều lắm . Mik đang rất cần
Do pt có 2 nghiệm \(x_1,x_2\) nên ta có :
\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{2}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)
Ta có :
\(P=x_1\left(3+x_2\right)+x_2\left(3+x_1\right)+3x^2_1+3x^2_2-10\)
\(=3x_1+x_1x_2+3x_2+x_1x_2+3\left(x_1^2+x_2^2\right)-10\)
\(=3\left(x_1+x_2\right)+2x_1x_2+3\left(x^2_1+x^2_2\right)-10\)
\(=3S+2P+3\left(S^2-2P\right)-10\)
\(=3.\left(-\dfrac{5}{2}\right)+2.\left(-\dfrac{1}{2}\right)+3\left(\left(-\dfrac{5}{2}\right)^2-2\left(-\dfrac{1}{2}\right)\right)-10\)
\(=\dfrac{13}{4}\)
Vậy \(P=\dfrac{13}{4}\)
\(a,2x^2+3x-9=0\\ \Leftrightarrow\left(2x^2+6x\right)-\left(3x+9\right)=0\\ \Leftrightarrow2x\left(x+3\right)-3\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(b,6x-12x^2=0\\ \Leftrightarrow6x\left(1-2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(c,8x^2-1=0\\ \Leftrightarrow x^2=\dfrac{1}{8}\\ \Leftrightarrow x=\pm\dfrac{\sqrt{2}}{4}\)
\(d,x^4-7x^2-18=0\\ \Leftrightarrow\left(x^4-3x^3\right)+\left(3x^3-9x^2\right)+\left(2x^2-6x\right)+\left(6x-18\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+3x^2+2x+6\right)=0\\ \Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+2\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2=-2\left(vô.lí\right)\end{matrix}\right.\)