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\(1a.\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}=21-2\sqrt{21}+2\sqrt{21}=21\) \(b.\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}=11+2\sqrt{30}-2\sqrt{30}=11\)
\(2a.\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}.b^2}+\sqrt{\dfrac{a^2}{b^2}.\dfrac{b}{a}}=\sqrt{\dfrac{a}{b}}+b\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{a}{b}}=\left(2+b\right)\sqrt{\dfrac{a}{b}}\) \(b.\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(x-1\right)^2}}.\sqrt{\dfrac{\left(2\sqrt{m}x-2\sqrt{m}\right)^2}{81}}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{\text{|}2\sqrt{m}x-2\sqrt{m}\text{|}}{9}=\dfrac{\sqrt{m}}{\text{|}x-1\text{|}}.\dfrac{2\sqrt{m}\text{|}x-1\text{|}}{9}=\dfrac{2m}{9}\) \(3a.VP=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1=VT\)
KL : Vậy đẳng thức được chứng minh.
\(b.VP=\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{\text{|}a+b\text{|}}=\dfrac{a+b}{b^2}.\dfrac{b^2\text{|}a\text{|}}{a+b}=\text{|}a\text{|}=VT\)
KL : Vậy đẳng thức được chứng minh .
P/s : Dài v ~
\(\Delta=\left(4m+3\right)^2-8\left(2m^2-1\right)=24m+17\)
a/ Để pt có 2 nghiệm pb \(\Leftrightarrow\Delta>0\Rightarrow24m+17>0\Rightarrow m>-\frac{17}{24}\)
b/ Pt có nghiệm kép \(\Leftrightarrow\Delta=0\Rightarrow m=-\frac{17}{24}\)
Nghiệm kép: \(x=-\frac{b}{2a}=\frac{4m+3}{4}=\frac{-\frac{17}{24}.4+3}{4}=\frac{1}{24}\)
c/ Pt vô nghiệm \(\Leftrightarrow\Delta< 0\Rightarrow m< -\frac{17}{24}\)
a: \(A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+2\sqrt{x}}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}+1+x-\sqrt{x}+2\sqrt{x}}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2\sqrt{x}+1}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)+2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b: Để A<0 thì \(\sqrt{x}-1< 0\)
=>0<x<1
3) \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\left(\dfrac{a+b+c}{abc}\right)=4-2=2\)
1) \(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^4+y^4+x^2y^2=21\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
Đặt \(\left(x^2+y^2;xy\right)=\left(a;b\right)\)
\(\left\{{}\begin{matrix}a+b=7\\a^2-b^2=21\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+b=7\\\left(a-b\right)\left(a+b\right)=21\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a+b=7\\a-b=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
Tới đây tiếp tục thay vào giải, lười rồi :D
bạn ơi -2 nha
a/ \( (a+1)x^2−2(a+3)x+2 =0\) (1)
Với \(a+1=0\Leftrightarrow a=-1\) thì \(\left(1\right)\Leftrightarrow-4x+2=0\Leftrightarrow x=\frac{1}{2}\)
Với \(a\ne-1\), ta có: \(\Delta'=\left(a+3\right)^2-2\left(a+1\right)=a^2+4a+7=\left(a+2\right)^2+3>0\forall x\in R\)
Suy ra ĐPCM
b/ \(x^ 2 +(a+1)x+2(a^ 2 −a+1) =0\)
có \(\Delta=\left(a+1\right)^2-4.2\left(a^2-a+1\right)=-7a^2+10a-7\)
Đề sai bạn nhé, vì phương trình có thể vô nghiệm nha bạn!