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Bài 1:
a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)
\(=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\)
\(=\sqrt{2}\)
b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2-\sqrt{5}\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2\)
Bài 2:
a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne\pm2\)
Với \(x\ne\pm2\), ta có:
\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)
\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)
\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)
\(\Rightarrow x^2-4=8-x\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\) (TM)
Vậy phương trình có tập nghiệm là: S ={3; -4}
\(x^2-4x-6=0\)
\(\text{Δ}=\left(-4\right)^2-4\cdot1\cdot\left(-6\right)=16+24=40>0\)
=>Phương trình này có hai nghiệm phân biệt
Theo vi-et, ta có:
\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-6}{1}=-6\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=4^2-2\cdot\left(-6\right)=16+12=28\)
\(B=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1\cdot x_2}=\dfrac{4}{-6}=-\dfrac{2}{3}\)
\(C=x_1^3+x_2^3\)
\(=\left(x_1+x_2\right)^3-3\cdot x_1\cdot x_2\cdot\left(x_1+x_2\right)\)
\(=4^3-3\cdot4\cdot\left(-6\right)=64+72=136\)
\(D=\left|x_1-x_2\right|\)
\(=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{4^2-4\cdot\left(-6\right)}=\sqrt{16+24}=\sqrt{40}=2\sqrt{10}\)
a. thay m=-4 vào (1) ta có:
\(x^2-5x-6=0\)
Δ=b\(^2\)-4ac= (-5)\(^2\) - 4.1.(-6)= 25 + 24= 49 > 0
\(\sqrt{\Delta}=\sqrt{49}=7\)
x\(_1\)=\(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+7}{2}\)=6
x\(_2\)=\(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-7}{2}\)=-1
vậy khi x=-4 thì pt đã cho có 2 nghiệm x\(_1\)=6; x\(_2\)=-1
Lời giải
a) Thay a=2+√3a=2+3 và b=2−√3b=2−3 vào P, ta được:
P=a+b−abP=2+√3+2−√3−(2+√3)(2−√3)P=2+2−(22−√32)P=4−(4−3)P=4−4+3=3P=a+b−abP=2+3+2−3−(2+3)(2−3)P=2+2−(22−32)P=4−(4−3)P=4−4+3=3
b) {3x+y=5x−2y=−3⇔{6x+2y=10x−2y=−3⇔{7x=7x−2y=−3⇔{x=1y=2{3x+y=5x−2y=−3⇔{6x+2y=10x−2y=−3⇔{7x=7x−2y=−3⇔{x=1y=2
Vậy nghiệm hệ phương trình (1; 2)
Có gì bạn tham khảo nha//
2)
\(A=\dfrac{5\sqrt{a}-3}{\sqrt{a}-2}+\dfrac{3\sqrt{a}+1}{\sqrt{a}+2}-\dfrac{a^2+2\sqrt{a}+8}{a-4}\)
\(=\dfrac{\left(5\sqrt{a}-3\right)\left(\sqrt{a}+2\right)+\left(3\sqrt{a}+1\right)\left(\sqrt{a}-2\right)-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}+\sqrt{a}-2-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{-a^2+8a-16}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{-\left(a-4\right)^2}{a-4}=4-a\)
1: Ta có: \(\left\{{}\begin{matrix}3x-y=2m-1\\x+y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=5m+1\\x+y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=3m+2-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=\dfrac{12m+8-5m-1}{4}=\dfrac{7m+7}{4}\end{matrix}\right.\)
Ta có: \(x^2+2y^2=9\)
\(\Leftrightarrow\left(\dfrac{5m+1}{4}\right)^2+2\cdot\left(\dfrac{7m+7}{4}\right)^2=9\)
\(\Leftrightarrow\dfrac{25m^2+10m+1}{16}+\dfrac{2\cdot\left(49m^2+98m+49\right)}{16}=9\)
\(\Leftrightarrow25m^2+10m+1+98m^2+196m+98-144=0\)
\(\Leftrightarrow123m^2+206m-45=0\)
Đến đây bạn tự làm nhé, chỉ cần giải phương trình bậc hai bằng delta thôi
Để (1) có 2 nghiệm dương \(\Rightarrow\left\{{}\begin{matrix}\Delta'=\left(m+3\right)^2-m-1\ge0\\x_1+x_2=2\left(m+3\right)>0\\x_1x_2=m+1>0\end{matrix}\right.\) \(\Rightarrow m>-1\)
\(P=\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|>0\Rightarrow P^2=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}\)
\(P^2=\dfrac{x_1+x_2-2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{2\left(m+3\right)-2\sqrt{m+1}}{m+1}=\dfrac{4}{m+1}-\dfrac{2}{\sqrt{m+1}}+2\)
\(P^2=\left(\dfrac{2}{\sqrt{m+1}}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\Rightarrow P\ge\dfrac{\sqrt{7}}{2}\)
Dấu "=" xảy ra khi \(\sqrt{m+1}=4\Rightarrow m=15\)
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)
\(P=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(\dfrac{x_1-x_2}{2}-\dfrac{x_1-x_2}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}-\dfrac{1}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}+1\right)^2\)
\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)
\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)
Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)