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a) Ta có:
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=\left(x+11\right)\left(\frac{1}{15}+\frac{1}{16}\right)\)
Mà ta có:
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\ne\frac{1}{15}+\frac{1}{16}\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Ta có:
\(A=1+x+x^2+x^3+...+x^{100}\)
Đặt \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
\(A=\left|x-13\right|+\left|x-14\right|+\left|x-15\right|+\left|x-16\right|+\left|x-17\right|-10\)
\(=\left(\left|x-13\right|+\left|x-16\right|\right)+\left(\left|x-14\right|+\left|x-17\right|\right)-10+\left|x-15\right|\)
\(=\left(\left|x-13\right|+\left|16-x\right|\right)+\left(\left|x-14\right|+\left|17-x\right|\right)-10+\left|x-15\right|\)
\(\Rightarrow A\ge\left|x-13+16-x\right|+\left|x-14+17-x\right|-10+\left|x-15\right|\)
\(=\left|3\right|+\left|3\right|-10+\left|x-15\right|\)\(=3+3-10+\left|x-15\right|=-6+\left|x-15\right|\)
Vì \(\left|x-15\right|\ge0\forall x\)\(\Rightarrow A\ge-6\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-13\right)\left(16-x\right)\ge0\\\left(x-14\right)\left(17-x\right)\ge0\\x-15=0\end{cases}}\Leftrightarrow\hept{\begin{cases}13\le x\le16\\14\le x\le17\\x=15\end{cases}}\Leftrightarrow x=15\)
Vậy \(minA=-6\Leftrightarrow x=15\)
Ta có : \(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\frac{\left(\sqrt{5}+2\sqrt[3]{\sqrt{5}-2^{ }}\right)^3}{\sqrt{5}+3-\sqrt{5}}\) 2)3 trong căn bậc nhé mk ko vt đc ( ko bt giải thick thông cảm )
\(=\frac{\sqrt{5}^2-2^2}{3}\)
\(=\frac{1}{3}\)
Vậy \(A=\left(3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2\right)^{2011}=3^{2011}\)
Trả lời
A=(3x3+8x2+2)2011 với x=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}\sqrt{9-6\sqrt{5}+5}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)
=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}\)
=\(\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{3}\)
=1/3
Học tốt !
(3x - 1)^2016 + (5y - 3)^2016 < 0 (1)
có (3x - 1)^2016 > 0
(5y - 3)^2018 > 0
=> (3x-1)^2016 + (5y - 3)^2018 > 0 và (1)
=> (3x - 1)^2016 + (5y - 3)^2016 = 0
=> 3x - 1 = 0 và 5y - 3 = 0
=> x = 1/23 và y = 3/5
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
Nếu x = 2014
Thì \(\left(x+14\right)^2.\left(x-14\right)\)= \(\left(2014+14\right)^2.\left(2014-14\right)\)
Sử dụng hằng đẳng thức ta được:
\(\left(2014^2-14^2\right).\left(2014+14\right)\)
\(\Leftrightarrow\left(4056196-196\right).2028\)
\(\Leftrightarrow4056000.2028\)
\(\Leftrightarrow8225568000\)
Chúc bạn học tốt !!!
Thay x=2014 vào biều thức ta được
A = \(\left(2014+14\right)^2\cdot\left(2014-14\right)\)
= \(2028\cdot2028\cdot2000\)
=8225568000
Chúc bạn hok tốt