Tìm x
│x^2 + │x-1│^2=x^2+2
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27 tháng 10 2021
Bài 1:
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{2x}{x-1}\)
29 tháng 8 2021
a: Ta có: \(M=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
b: Để M>1 thì M-1>0
\(\Leftrightarrow\dfrac{x^2-x+1}{x-1}>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
31 tháng 7 2023
a: \(Y=\dfrac{3\left(x^2-x-1\right)-x^2+1}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{1-1+x}{1-x}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{-x}{x-1}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}-\dfrac{x-2}{x-1}\)
\(=\dfrac{2x^2-3x-2-x^2+4}{\left(x+2\right)\left(x-1\right)}=\dfrac{x^2-3x+2}{\left(x+2\right)\left(x-1\right)}=\dfrac{x-2}{x+2}\)
b: Y=2
=>2x+4=x-2
=>x=-6(nhận)
c; Y nguyên
=>x+2-4 chia hết cho x+2
=>x+2 thuộc {1;-1;2;-2;4;-4}
Kết hợp ĐKXĐ, ta được: x thuộc {-1;-3;-4;-6}
16 tháng 6 2023
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |2x+1|=3
=>x=1(loại); x=-2(nhận)
Khi x=-2 thì P=4/-3=-4/3
c: P=-1/2
=>x^2/x-1=-1/2
=>2x^2=-x+1
=>2x^2+x-1=0
=>2x^2+2x-x-1=0
=>(x+1)(2x-1)=0
=>x=1/2; x=-1
19 tháng 5 2022
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
19 tháng 5 2022
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
ND
Nguyễn Đức Trí
VIP
11 tháng 8 2023
\(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(1\right)\)
a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)
\(\left(1\right)\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x+1}\)
b) Để \(A=-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x^2}{x+1}=-\dfrac{1}{2}\left(x\ne-1\right)\)
\(\Leftrightarrow2x^2=-\left(x+1\right)\)
\(\Leftrightarrow2x^2+x+1=0\)
\(\Delta=1-8=-7< 0\)
Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)
c) Để \(A< 1\)
\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)
\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)
\(\Leftrightarrow x^2-x-1< 0\)
\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)
\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)
\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)
d) Để A nguyên
\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)
\(\Leftrightarrow x^2⋮x+1\)
\(\Leftrightarrow x^2-x\left(x+1\right)⋮x+1\)
\(\Leftrightarrow x^2-x^2+x⋮x+1\)
\(\Leftrightarrow x⋮x+1\)
\(\Leftrightarrow x-x-1⋮x+1\)
\(\Leftrightarrow-1⋮x+1\)
\(\Leftrightarrow x+1\in\left\{-1;1\right\}\)
\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)
2 tháng 11 2017
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3