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Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
Áp dụng BĐT AM-GM: \(\dfrac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\le\dfrac{1}{4}\left(4a+4b\right)=a+b\)
Ta chứng minh: \(3\left(a+b\right)^2+4ab\ge2\left(a+b\right)\)
hay \(3\left(a+b\right)^2+4ab\ge2\left(a+b\right)\left(\sqrt{a}+\sqrt{b}\right)^2\)
\(\Leftrightarrow\left(a+b-2\sqrt{ab}\right)^2\ge0\)( đúng)
Dấu = xảy ra khi \(a=b=\dfrac{1}{4}\)
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0
khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)
thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay
\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)
Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)
Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)
Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)
Vậy \(x+y=1;xy=\dfrac{1}{4}\)
Do đó x,y là nghiệm của phương trình:
\(t^2-t+\dfrac{1}{4}=0\)
\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)
Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)
Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
mk lm 1 bài còn lại bn lm tương tự nha :
a) điều kiện xác định : \(x\ge0;y\ge1\)
đặc \(a=\sqrt{x};b=\sqrt{y-1}\)
\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}a+2b=5\\4a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
ta có : \(a=1\Rightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tmđk\right)\) ; \(b=2\Rightarrow\sqrt{y-1}=2\Leftrightarrow y=5\left(tmđk\right)\)
vậy phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;5\right)\)
b) bn đặc : \(a=\dfrac{1}{x};b=\dfrac{1}{y+12}\)
c) bn đặc : \(a=\dfrac{x}{x+1};b=\dfrac{y}{y+1}\)
nhớ điều kiện nha
áp dụng bất đẳng thức mincopski ta có :
\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}=\sqrt{3^2+\left(\dfrac{9}{3}\right)^2}=3\sqrt{2}\)
\(\Rightarrow GTNN\) của \(S\) là \(3\sqrt{2}\) dấu "=" xảy ra khi \(a=b=c=1\)
\(P=\dfrac{4}{2\sqrt{a+3}}+\dfrac{2}{2\sqrt{b+3}}\ge\dfrac{4}{\dfrac{a+3+4}{2}}+\dfrac{2}{\dfrac{b+3+4}{2}}\)
\(\Rightarrow P\ge\dfrac{16}{2a+14}+\dfrac{4}{b+7}\ge\dfrac{\left(4+2\right)^2}{2a+b+21}\ge\dfrac{36}{3+21}=\dfrac{3}{2}\)
\(\Rightarrow P_{min}=\dfrac{3}{2}\) khi \(a=b=1\)
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+ac+bc}{abc}=0\Leftrightarrow ab+ac+bc=0\)
Vì a,b>0\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}>0\)
Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Suy ra \(\dfrac{1}{c}< 0\Leftrightarrow c< 0\)
\(\Leftrightarrow c+\left|c\right|=0\Leftrightarrow c+\sqrt{c^2}=0\Leftrightarrow c+\sqrt{ab+ac+bc+c^2}=0\)(vì ab+ac+bc=0)\(\Leftrightarrow c+\sqrt{a\left(b+c\right)+c\left(b+c\right)}=0\Leftrightarrow c+\sqrt{\left(b+c\right)\left(a+c\right)}=0\Leftrightarrow2c+2\sqrt{\left(b+c\right)\left(a+c\right)}=0\Leftrightarrow a+b=a+b+2c+2\sqrt{\left(b+c\right)\left(a+c\right)}\Leftrightarrow a+b=\left(b+c\right)+2\sqrt{\left(b+c\right)\left(a+c\right)}+\left(a+c\right)\Leftrightarrow a+b=\left(\sqrt{b+c}+\sqrt{a+c}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{b+c}+\sqrt{a+c}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{b+c}+\sqrt{a+c}\)
\(\dfrac{1}{a-b}-\dfrac{1}{b}=a+b+\dfrac{a}{b}+\dfrac{b}{a}+1\)
\(\Leftrightarrow1=\left(a-b\right)\left(a+b+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{1}{b}+1\right)\circledast\)
VP của \(\circledast\Leftrightarrow a^2+ab+\dfrac{a^2}{b}+b+\dfrac{a}{b}+a-ab-b^2-a-\dfrac{b^2}{a}-1-b=a^2-b^2+\dfrac{a^2}{b}-\dfrac{b^2}{a}+\dfrac{a}{b}-1\)
Do : \(a^2=2;b^3=2;\dfrac{a^2}{b}=\dfrac{2}{b}=b^2;\dfrac{a}{b}=\dfrac{b^2}{a}\)
\(\Rightarrow2-\dfrac{2}{b}+\dfrac{2}{b}-\dfrac{a}{b}+\dfrac{a}{b}-1=1=VT\)
=> đpcm