\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}+\dfrac{1}{x-20}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{x-1}=\dfrac{3}{4}\Rightarrow3x-3=4\Rightarrow x=\dfrac{7}{3}\)

Vậy...

1)Ta có: M=1+2+2²+…+2²⁰⁶

=>M=1+(2+2²+…+2²⁰⁶)

=>M=1+2.(1+2+…+2²⁰⁵)

Vì 2.(1+2+…+2²⁰⁵) chia hết cho 2

=>1+2.(1+2+…+2²⁰⁵) không chia hết cho 2

=>M không chia hết cho 2

\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{a^3}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)

\(=a+\sqrt{ab}+b\)

121 nhé bạn

Tick tớ đc chứ 

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

a, Ta có: \(B=2x^2+10x-1=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)

\(=2\left(x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}\right)-\dfrac{27}{2}\)

\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge\dfrac{-27}{2}\)

Dấu " = " khi \(2\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)

Vậy \(MIN_B=\dfrac{-27}{2}\) khi \(x=\dfrac{-5}{2}\)

b, Ta có: \(C=5x-x^2=-\left(x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

Dấu " = " khi \(-\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(MAX_C=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

Ta có: \(\left|x-y\right|+\left|x-1\right|\ge0\)

\(\Rightarrow A=\left|x-y\right|+\left|x-1\right|+2017\ge2017\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|x-1\right|=0\end{matrix}\right.\Rightarrow x=y=1\)

Vậy \(MIN_A=2017\) khi x = y = 1

Áp dụng bất đẳng thức Cauchy-Shwarz dạng Engel ta có:

\(A^2=2\left(x^2+y^2\right)\ge2.\dfrac{\left(x+y\right)^2}{2}=1^2=1\)

\(\Rightarrow A\ge1\) ( do \(A\ge0\) )

Dấu " = " khi \(x=y=\dfrac{1}{2}\)

Vậy \(MIN_A=1\) khi \(x=y=\dfrac{1}{2}\)

bài này chuyển vế bình thường, sau khi đó thì xét các trường hợp của x để phá dấu giá trị tuyệt đối

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{98}{99}.\dfrac{100}{99}.\dfrac{99}{100}.\dfrac{101}{100}\)

\(=\dfrac{1.2...98.99}{2.3...99.100}.\dfrac{3.4...99.101}{2.3...99.100}\)

\(=\dfrac{1}{100}.\dfrac{101}{2}\)

\(=\dfrac{101}{200}\)