23. Cho 2 bt
A = \(\dfrac{1}{x-2}\) và B = \(\dfrac{x^2-x-2}{x^2-10x+10}\)\(-\dfrac{2x-4}{x-5}\)( x khác 2 ; x khác 5)
a. Đặt P = A + B . Rút gọn P
b. Tìm x nguyên để P đạt giá trị nguyên lớn nhất
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5 tháng 7 2022
`a)`\(M=A.B\)
\(M=\left(\dfrac{x+2}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)
\(M=\left(\dfrac{-\left(x+2\right)^2-4x^2-\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-3}{2x-x^2}\right)\)
\(M=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)
\(M=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}.\dfrac{2x-x^2}{x-3}\)
\(M=\dfrac{-4x\left(2x-x^2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(M=\dfrac{4x^2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(M=\dfrac{4x^2}{x-3}\)
`b)`\(\left|x-7\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)
`@` Với `x=11`\(\Rightarrow M=\dfrac{121}{2}\)
`@` Với `x=3` `=>` `M` không xác định
`c)`\(A>0\)
\(\Leftrightarrow\dfrac{4x^2}{x-3}>0\)
`<=>x-3>0`
`<=>x>3`
5 tháng 7 2022
`a)`\(S=1+A:B\)
\(S=1+\left(\dfrac{x^3-2x^2}{x^3-x^2+x}\right):\left(\dfrac{x+1}{x^3+1}+\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right)\)
\(S=1+\left(\dfrac{x^2-2x}{x^2-x+1}\right):\left(\dfrac{x+1+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)
\(S=1+\dfrac{x^2-2x}{x^2-x+1}:\dfrac{4x-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(S=1+\dfrac{x^2-2x}{x^2-x+1}.-\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{2\left(x^2-2x\right)}\)
\(S=1-\dfrac{x+1}{2}\)
`b)`\(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=-\dfrac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`@`Với `x=2` \(\Rightarrow S=1+\dfrac{2+1}{2}=1+\dfrac{3}{2}=\dfrac{5}{2}\)
`@`Với `x=-1/2` \(\Rightarrow S=1+\dfrac{-\dfrac{1}{2}+1}{2}=\dfrac{5}{4}\)
`c)`\(S=1-\dfrac{x+1}{2}\)
Để `S` nguyên thì \(x+1⋮2\) hay `x` thuộc các số lẻ
5 tháng 7 2022
`a)`\(P=A:B\)
\(P=\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\)
\(P=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(P=\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-x^2}{\left(x-1\right)\left(x+1\right)}\)
\(P=-\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)
`b)`\(P=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(2x^2+2\right)=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow4x^2+4=x^2-1\)
\(\Leftrightarrow3x^2=-5\) ( vô lý )
Vậy không có giá trị `x` thỏa mãn `P=1/2`
ND
0
NM
0
4 tháng 7 2022
We subtitute \(ab+bc+ca=1\) into \(a^2+1\). We have: \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)\)\(=\left(a+b\right)\left(a+c\right)\)
Similarly, we have \(b^2+1=\left(a+b\right)\left(b+c\right)\) and \(c^2+1=\left(a+c\right)\left(b+c\right)\)
From these, we have \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)\(=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Thus, we must have \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)\(=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)
Because both \(a,b,c\) are rational numbers, \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) must be a rational number. Therefore, \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\) is also a rational number.
4 tháng 7 2022
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)
\(=\sqrt{\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)}\)
\(=\sqrt{\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(b+c\right)+a\left(b+c\right)\right]\left[c\left(c+a\right)+b\left(c+a\right)\right]}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}\)
\(=\sqrt{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)
Do \(a,b,c\) là các số hữu tỉ nên \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) là số hữu tỉ.
\(\Rightarrowđpcm\)
4 tháng 7 2022
\(a,b\ne0\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{b^2\left(a+b\right)^2+a^2\left(a+b\right)^2+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2b^2+2ab^3+b^4\right)+\left(a^4+2a^3b+a^2b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^4+2a^2b^2+b^4\right)+2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+b^2+ab\right)^2}{a^2b^2\left(a+b\right)^2}}=\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\)
Do \(a,b\) là số hữu tỉ nên \(\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\) cũng là số hữu tỉ.
\(\Rightarrowđpcm\)