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a,\(P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
\(b,\) Để \(P=-\dfrac{1}{2}\) hay \(\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
\(\Leftrightarrow2x^2=-\left(x-1\right)\)
\(\Leftrightarrow2x^2=-x+1\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow2x^2+2x-x-1=0\)
\(\Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
a) ĐKXĐ : để phân thức P xác định :
x2 - 2x + 1 ≠ 0 -> x ≠ 1
x ≠ 0
x - 1 ≠ 0 -> x≠ 1
x2 - x ≠ 0 -> x ≠ 0 ;1\
a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
\(P=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: Để \(P=-\dfrac{1}{2}\) thì \(\dfrac{x^2}{x-1}=\dfrac{-1}{2}\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\)
=>x=-1(loại)hoặc x=1/2(nhận)
a/ (1+x2).(1+x)
b/A=\(\dfrac{-68}{27}\)
c/x>-1 và x2 >1
phần giải tự lm nhé
Lời giải:
a) ĐKXĐ: \(x\neq \pm 1\)
Ta có: \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\left(\frac{(1-x)(1+x+x^2)}{1-x}-x\right): \frac{1-x^2}{(1-x)-x^2(1-x)}\)
\(=(1+x+x^2-x):\frac{1-x^2}{(1-x)(1-x^2)}=(1+x^2):\frac{1}{1-x}=(x^2+1)(1-x)\)
b) Tại \(x=-1\frac{2}{3}=\frac{-5}{3}\Rightarrow A=(\frac{25}{9}+1)(1-\frac{-5}{3})=\frac{272}{27}\)
c) Để \(A=(x^2+1)(1-x)>0\)
\(\Rightarrow 1-x>0\) (do \(x^2+1>0\) )
\(\Rightarrow x< 1\)
Vậy \(x<1; x\neq -1\)
Bài 1:
Áp dụng BĐT Cô-si cho các số dương ta có:
\(x+2017\geq 2\sqrt{x.2017}\Rightarrow (x+2017)^2\geq 8068x\)
\(\Rightarrow M=\frac{x}{(x+2017)^2}\leq \frac{x}{8068x}=\frac{1}{8068}\)
Vậy GTLN của \(M=\frac{1}{8068}\)
Dấu "=" xảy ra khi $x=2017$
Bài 2:
Thay $y=1-x$ vào biểu thức $M$ ta có:
\(M=5x^2+y^2=5x^2+(1-x)^2\)
\(=5x^2+(x^2-2x+1)=6x^2-2x+1\)
\(=6(x^2-\frac{1}{3}x+\frac{1}{36})+\frac{5}{6}\)
\(=6(x-\frac{1}{6})^2+\frac{5}{6}\geq 6.0+\frac{5}{6}=\frac{5}{6}\)
Vậy GTNN của $M$ bẳng $\frac{5}{6}$ khi \(x=\frac{1}{6}; y=\frac{5}{6}\)
2, Ta có: A= \(\left(1+\dfrac{1}{x}\right)^2+\left(1+\dfrac{1}{y}\right)^2=1+\dfrac{2}{x}+\dfrac{1}{x^2}+1+\dfrac{2}{y}+\dfrac{1}{y^2}\)
\(=2+2\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{x^2}+\dfrac{1}{y^2}=2+2.\dfrac{x+y}{xy}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\)
\(=2+2.\dfrac{1}{xy}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\) ( do x+y=1)
Ta cm được BĐT : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với a, b >0
Áp dụng BĐT ta được: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{x^2+y^2}\) ( do x, y >0)
=> \(A=2+2.\dfrac{1}{xy}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2+2.\dfrac{1}{xy}+\dfrac{4}{x^2+y^2}=2+\dfrac{4}{2xy}+\dfrac{4}{x^2+y^2}\)
Áp dụng BĐT ta được: \(\dfrac{4}{2xy}+\dfrac{4}{x^2+y^2}\ge\dfrac{16}{2xy+x^2+y^2}=\dfrac{16}{\left(x+y\right)^2}=\dfrac{16}{1}=16\) ( do x+y=1)
=> \(A\ge2+\dfrac{4}{2xy}+\dfrac{4}{x^2+y^2}\ge2+16=18\)
dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
vậy GTNN của A = 18 khi \(x=y=\dfrac{1}{2}\)
1a) x2 - 5 > 4
<=> x2 - 9 > 0
<=> ( x - 3)( x + 3) > 0
x x-3 x+3 -3 3 0 0 - - + - + + (x-3)(x+3) + 0 - 0 + Vậy , để : x2 - 9 > 0 thì : x < - 3 hoặc x > 3
b) tương tự nhé
2. \(\dfrac{x+6}{5}-\dfrac{x-2}{3}\) ≥ 2
<=> \(\dfrac{3\left(x+6\right)-5\left(x-2\right)}{15}\) ≥ \(\dfrac{30}{15}\)
<=> 3x + 18 - 5x + 10 ≥ 30
<=> 28 - 2x ≥ 30
<=> 2x ≤ -2
<=> x ≤ -1
KL....
3. ( x + 3 )( 1 - x) ≤ 0
Lập bảng xét dấu :
x x+3 1-x (x+3)(1-x) -3 1 0 0 0 0 - + + + + - - + -
Nhìn bảng xét dấu ta thấy : x ≤ - 3 hoặc : x ≥ 1 ( vô lý )
Vậy, BPT vô nghiệm
a: \(M=1:\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=1:\dfrac{x^2-x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}=\dfrac{x^2+x+1}{x}\)
b: \(M-3=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}>0\)
=>M>3
e: Khi x=1/4 thì \(M=\dfrac{\dfrac{1}{16}+\dfrac{1}{4}+1}{\dfrac{1}{4}}=\dfrac{21}{4}\)
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
\(A=x+1+\dfrac{1}{x-1}\\ \\ =x-1+2+\dfrac{1}{x-1}\\ =\left(x-1\right)+\dfrac{1}{x-1}+2\)
Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow A=\left(x-1\right)+\dfrac{1}{x-1}+2\ge2+2\ge4\)
Dấu "=" xảy ra khi:
\(x-1=1\\ \Leftrightarrow x=2\)
Vậy \(A_{Min}=4\) khi \(x=2\)