cho \(A=\frac{4}{x+2};B=\frac{x+1}{2}\)
a, tìm x để A.B=0
b, Tìm x nguyên để A.B là 1 số nguyên
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22 tháng 5 2021
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
EK
25 tháng 5 2021
Bạn có thể viết thay dòng "Từ (1) và (2)" thành "Từ các điều kiện trên" bạn nhé !(bạn ko cần phải sửa, đây chỉ là gợi ý)
CV
1 tháng 8 2017
Ngọc Anh Dũngo0oNguyễno0oHuy hoàng indonaca0o0 khùng mà 0o0Tình bạn vĩnh cửu Phương DungHacker Mũ Trắng
2 tháng 6 2015
Đặt \(t=\frac{x}{y}+\frac{y}{x}\). Vì x; y > 0 => \(\frac{x}{y}>0;\frac{y}{x}>0\). Áp dung BDT Cô - si có:
\(t=\frac{x}{y}+\frac{y}{x}\ge2.\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)
Có: \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=\left(\frac{x}{y}+\frac{y}{x}\right)^2-2.\frac{x}{y}.\frac{y}{x}=t^2-2\)
\(\frac{x^4}{y^4}+\frac{y^4}{x^4}=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)^2-2.\frac{x^2}{y^2}.\frac{y^2}{x^2}=\left(t^2-2\right)^2-2=t^4-4t^2+4-2=t^4-4t^2+2\)
Vậy \(A=t^4-4t^2+2-\left(t^2-2\right)+t=t^4-5t^2+t+4\)
=> \(A=\left(t^4-8t^2+16\right)+3t^2+t-12=\left(t^2-4\right)^2+3t^2+t-12=\left(t^2-4\right)^2+3\left(t^2-4\right)+t\ge2\)với mọi \(t\ge2\)
Vì \(t\ge2\) => \(t^2\ge4\Rightarrow t^2-4\ge0\)
Vậy Min A = 2 khi t = 2 <=> \(\frac{x}{y}+\frac{y}{x}=2\) <=> x = y = 1
HN
3 tháng 11 2019
Ta có:
\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)
Chúc bạn học tốt!
9 tháng 8 2019
tổng 2 số là 150, tổng của 1/6 số này và 1/9 số kia = 18. Tìm 2 số đó
14 tháng 7 2016
a) ĐKXĐ : \(0\le x\ne4\)
b) \(A=\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\left(x-4\right)\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(=\frac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=-1\)
13 tháng 7 2016
\(A=\left[\frac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{x-4}\right]:\frac{1}{x-4}\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{x-4}.\left(x-4\right)\)=\(=\frac{-1}{x-4}.\left(x-4\right)=-1\)
Vậy giá trị của A thỏa mãn mọi x và rút gọn lại còn -1
22 tháng 7 2020
Bài 1:
a) Ta có: \(P=\frac{x}{x+2}+\frac{x+3}{x-2}+\frac{6-9x}{4-x^2}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6-9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x+x^2+5x+6-6+9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Để P=3 thì \(\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}=3\)
\(\Leftrightarrow2x^2+12x=3\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow2x^2+12x=3\left(x^2-4\right)\)
\(\Leftrightarrow2x^2+12x=3x^2-12\)
\(\Leftrightarrow2x^2+12x-3x^2+12=0\)
\(\Leftrightarrow-x^2+12x+12=0\)
\(\Leftrightarrow x^2-12x-12=0\)
\(\Leftrightarrow x^2-12x+36-24=0\)
\(\Leftrightarrow\left(x-6\right)^2=24\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=\sqrt{24}\\x-6=-\sqrt{24}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6+2\sqrt{6}\left(nhận\right)\\x=6-2\sqrt{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: khi P=3 thì \(x\in\left\{6+2\sqrt{6};6-2\sqrt{6}\right\}\)
Bài 2:
a) Ta có: \(B=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)
\(=\frac{2a^2}{\left(a+1\right)\left(a-1\right)}+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a^2+a^2-a-a^2-a}{\left(a+1\right)\cdot\left(a-1\right)}=\frac{2a^2-2a}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}=\frac{2a}{a+1}\)
b) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)
Để B là số nguyên thì \(2a⋮a+1\)
\(\Leftrightarrow2a+2-2⋮a+1\)
\(\Leftrightarrow-2⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(-2\right)\)
\(\Leftrightarrow a+1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{0;-2;1;-3\right\}\)
mà \(a\notin\left\{1;-1\right\}\)
nên \(a\in\left\{0;-2;-3\right\}\)
Vậy: khi B có giá trị nguyên thì \(a\in\left\{0;-2;-3\right\}\)
Bài 3:
Ta có: \(Q=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x-2}\)
Bài 4:
a) Ta có: \(P=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)
\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right)\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}+2+3\sqrt{x}-6}{\sqrt{x}-2}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\cdot4\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-2\right)^2}\)
\(=\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Để P=-4 thì \(\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}=-4\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(\sqrt{x}-2\right)^2\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(x-4\sqrt{x}+4\right)\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4x+16\sqrt{x}-16\)
\(\Leftrightarrow-16x+16\sqrt{x}+4x-16\sqrt{x}+16=0\)
\(\Leftrightarrow-12x+16=0\)
\(\Leftrightarrow-12x=-16\)
hay \(x=\frac{4}{3}\)(nhận)
Vậy: Khi P=-4 thì \(x=\frac{4}{3}\)
a) Vì A.B = 0(x khác -2)
\(\Rightarrow\frac{4}{x+2}.\frac{x+1}{2}=0\)
\(\Rightarrow\frac{2x+2}{x+2}=0\)
\(\Rightarrow2x+2=0\Leftrightarrow x=-1\)
b) A . B = \(\frac{2x+2}{x+2}=\frac{2x+4-2}{x+2}=2-\frac{2}{x+2}\)
Để A.B nguyên \(\Rightarrow x+2\inƯ\left(2\right)=\left(1;-1;2;-2\right)\)
\(\Rightarrow x\in\left(-1;-3;0;-4\right)\)