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\(\left(\dfrac{x-4}{2x-4}+\dfrac{2}{x^2-2x}\right):\dfrac{x-2}{x+1}\)
\(=\left(\dfrac{x-4}{2\left(x-2\right)}+\dfrac{2}{x\left(x-2\right)}\right).\dfrac{x+1}{x-2}\)
\(=\dfrac{x\left(x-4\right)+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{x^2-4x+4}{2x\left(x-2\right)}.\dfrac{x+1}{x-2}\)
\(=\dfrac{\left(x-2\right)^2\left(x+1\right)}{2x\left(x-2\right)\left(x-2\right)}\)
\(=\dfrac{x+1}{2x}\)
Mình làm nốt bài 2 nhé :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)
⇔ \(\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)
⇔ \(\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
⇔ \(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c\)
⇔ \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
Lời giải:
a)
Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)
Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)
\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)
b)
ĐK: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :
\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)
Vậy..........
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
B1:
a. \(\sqrt{\dfrac{4}{2x+3}}\)được xác định khi:\(\dfrac{4}{2x+3}\ge0\Leftrightarrow2x+3>0\Leftrightarrow x>-\dfrac{3}{2}\)
b.\(\sqrt{x\left(x+2\right)}\text{ }\) được xác định khi :\(x\left(x+2\right)\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\le-2\end{matrix}\right.\)
c.\(\sqrt{\dfrac{2x-1}{2-x}}\) được xác định khi :\(\dfrac{2x-1}{2-x}\ge0\Leftrightarrow\dfrac{1}{2}\le x< 2\)
B2:
a.\(\sqrt{\left(\sqrt{3}-2\right)^2}=|\sqrt{3}-2|=2-\sqrt{3}\) ( vì \(\sqrt{3}< \sqrt{4}=2\))
b.\(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=|\sqrt{3}-1|=\sqrt{3}-1\)(vì \(\sqrt{3}>\sqrt{1}=1\))
c.\(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=|\sqrt{5}-2|=\sqrt{5}-2\)(vì \(\sqrt{5}>\sqrt{4}=2\))
B3:
a.\(\sqrt{25-20x+4x^2}+2x=5\)
\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)
\(\Leftrightarrow|5-2x|+2x=5\) (1)
Nếu \(5-2x\le0\Leftrightarrow x\ge\dfrac{5}{2}\).Khi đó :
(1)\(\Leftrightarrow2x-5+2x=5\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)(thoả mãn đk)
Nếu \(5-2x>0\Leftrightarrow x< \dfrac{5}{2}\).Khi đó :
(1)\(\Leftrightarrow5-2x+2x=5\Leftrightarrow5=5\)(luôn đúng với mọi x )
kết hợp với điều kiện ta được :\(x< \dfrac{5}{2}\)
Vậy nghiệm của phương trình đã cho là \(x=\dfrac{5}{2}\) hoặc \(x< \dfrac{5}{2}\)
b.\(\sqrt{x^2+\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\)
\(\Leftrightarrow\sqrt{\left(x+\dfrac{1}{4}\right)^2}=\dfrac{1}{4}-x\)
\(\Leftrightarrow|x+\dfrac{1}{4}|=\dfrac{1}{4}-x\) (2)
Nếu \(x+\dfrac{1}{4}\le0\Leftrightarrow x\le-\dfrac{1}{4}\).Khi đó :
(2)\(\Leftrightarrow-\left(x+\dfrac{1}{4}\right)=\dfrac{1}{4}-x\Leftrightarrow\dfrac{1}{4}-x=\dfrac{1}{4}-x\) (luôn đúng với mọi x)
kết hợp với điều kiện ta được :\(x\le-\dfrac{1}{4}\)
Nếu \(x+\dfrac{1}{4}>0\Leftrightarrow x>-\dfrac{1}{4}\).Khi đó :
(2)\(\Leftrightarrow x+\dfrac{1}{4}=\dfrac{1}{4}-x\Leftrightarrow2x=0\Leftrightarrow x=0\)(tmđk)
Vậy nghiêm của phương trình là \(x\le-\dfrac{1}{4}\) hoặc \(x=0\)
c.\(\sqrt{x-2\sqrt{x-1}}=2\) (đkxđ :\(x\ge1\))
\(\Leftrightarrow\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow|\sqrt{x-1}-1|=2\)
\(\Leftrightarrow\sqrt{x-1}-1=2ho\text{ặc}\sqrt{x-1}-1=-2\)
\(\Leftrightarrow\sqrt{x-1}=3ho\text{ặc}\sqrt{x-1}=-1\)(vô nghiệm )
\(\Leftrightarrow x=10\)(tmđk )
Vậy nghiệm của phương trình đã cho là \(x=10\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
Lời giải:
a) ĐK: \(\left\{\begin{matrix} x-2\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow x-2>0\Leftrightarrow x>2\)
b) ĐK: \(\left\{\begin{matrix} x+2\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow x\geq 2\)
c) ĐK: \(\left\{\begin{matrix} x^2-4\neq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-2)(x+2)\neq 0\\ x\geq 2\end{matrix}\right.\Leftrightarrow x>2\)
d) ĐK: \(3-2x>0\Leftrightarrow x< \frac{3}{2}\)
e) ĐK: \(2x+3>0\Leftrightarrow x> \frac{-3}{2}\)
f) ĐK: \(x+1< 0\Leftrightarrow x< -1\)
Bài 1:
A.\(\left(\sqrt{x}+2\right)\) = -1 (ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{1}{x-4}\left(\sqrt{x}+2\right)=-1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-2}=-1\)
\(\Leftrightarrow\sqrt{x}-2=-1\)
\(\Leftrightarrow\sqrt{x}=1\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy x = 1
Bài 2: ĐK: \(x\ge0\)
Để \(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)\)\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3\right\}\)\(\Leftrightarrow x\in\left\{1\right\}\)
Bài 3:
a, Ta có: \(x+\sqrt{x}+1=x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}+1\\ =\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Ta có: 2 > 0 và \(x+\sqrt{x}+1>0\Rightarrow C>0\) và \(x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(C=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có: \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow x=0\left(TM\right)\)
Vậy MaxC = 2 khi x = 0
Còn cái GTNN chưa tính ra được, để sau nha
Bài 4: ĐK: \(x\ge0,x\ne1\)
\(D=\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\left(\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\sqrt{x}-1\)
\(D=3\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=2\left(TM\right)\)
\(D=x-3\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}-1=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(1-\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=9\left(TM\right)\end{matrix}\right.\)
Bài 5: \(E< -1\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{-3x}{2x+4\sqrt{x}}+1< 0\Leftrightarrow\dfrac{-3x+2x+4\sqrt{x}}{2x+4\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-x}{2x+4\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)
Ta có: \(\sqrt{x}>0\Leftrightarrow x>0\Leftrightarrow2x+4\sqrt{x}>0\) mà \(\dfrac{\sqrt{x}\left(4-\sqrt{x}\right)}{2x+4\sqrt{x}}< 0\)\(\Rightarrow\sqrt{x}\left(4-\sqrt{x}\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\left(L\right)\\4-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>0\\4-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\\left\{{}\begin{matrix}x>0\\x< 16\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 16,x\ne0\\0< x< 16\end{matrix}\right.\)
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
a/ \(N=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}\right).\left(-1\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{2^2-x^2}\right)\)
\(N=-\left(\dfrac{1}{x-2}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=-\left(-\dfrac{1}{2+x}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=-\left(\dfrac{-\left(2-x\right)-2x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-2+x-2x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-2-x}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(\dfrac{-\left(2+x\right)}{\left(2-x\right)\left(2+x\right)^2}\right)\)
\(N=-\left(-\dfrac{1}{\left(2-x\right)\left(2+x\right)}\right)\)
\(N=\dfrac{1}{4-x^2}\)
b/ ĐK của N: \(4-x^2\ne0\Leftrightarrow x\ne\pm2\)
\(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Tới đây bạn thay vào tính.
c/ Thay N = 1/2 vào:
\(\Rightarrow\dfrac{1}{2}=\dfrac{1}{4-x^2}\)
\(\Leftrightarrow4-x^2=2\)
\(\Leftrightarrow x^2=2\)
\(\Rightarrow x=\pm\sqrt{2}\left(loai\right)\)
a/ ĐKXĐ: \(x\ne\pm2\)
N= \(\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right).\left(\dfrac{2}{x}-1\right)\)
= \(\left(\dfrac{-1}{2-x}-\dfrac{2x}{\left(2-x\right)\left(2+x\right)}+\dfrac{1}{2+x}\right).\dfrac{2-x}{x}\)
= \(\dfrac{-\left(2+x\right)-2x+2-x}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x}\)
= \(\dfrac{-4x\left(2-x\right)}{x\left(2-x\right)\left(2+x\right)}\)
= \(\dfrac{-4}{2+x}\) (1)
b/ Ta có:
\(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{2}\end{matrix}\right.\) (t/m đk)
Thay x=0 vào (1) ta được:
\(N=\dfrac{-4}{2+0}=\dfrac{-4}{2}=-2\)
Vậy khi x=0 thì N=-2
Thay \(x=\dfrac{-1}{2}\) vào (1) ta được:
\(N=\dfrac{-4}{2-\dfrac{1}{2}}=\dfrac{-4}{\dfrac{3}{2}}=\dfrac{-8}{3}\)
Vậy khi \(x=\dfrac{-1}{2}\) thì \(N=\dfrac{-8}{3}\)
c/ Với \(x\ne\pm2\)
Để \(N=\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{-4}{2+x}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{-8}{2\left(2+x\right)}=\dfrac{2+x}{2\left(2+x\right)}\)
\(\Leftrightarrow-8-2-x=0\)
\(\Leftrightarrow x=-10\) (t/m đk)
Vậy để \(N=\dfrac{1}{2}\) thì x=-10