\(D=\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)

\(D=\frac{7}{4}\left(\frac{33}{12}+\frac{33.101}{20.101}+\frac{33.10101}{30.10101}+\frac{33.1010101}{42.1010101}\right)\)

\(D=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(D=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(D=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(D=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(D=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(D=\frac{7}{4}.33.\frac{4}{21}=\frac{7}{4}.3.11.\frac{4}{3.7}=11\)

BT1: 

ĐK: \(a>0,a\ne1\).

\(A=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{a+\sqrt{a}-2-\left(a-\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(A=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{2}{a-1}\)

ĐK: \(a\ge0,a\ne1\).

\(B=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(B=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(B=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

BT2: 

a) ĐK: \(a\ge0,a\ne1\).

\(A=\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\left(\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}\right).\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{a-\sqrt{a}-2-\left(a+\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(1-a\right)^2}{2}\)

\(A=\frac{-2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\left(\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)^2}{2}\)

\(A=-\sqrt{a}\left(\sqrt{a}-1\right)=\sqrt{a}-a\)

b) \(A=\sqrt{a}-a=\sqrt{a}\left(1-\sqrt{a}\right)>0\)

(Vì \(0< a< 1\Rightarrow1-\sqrt{a}>0\))

c) \(A=\sqrt{a}-a=\frac{1}{4}-\left(\frac{1}{4}-\sqrt{a}+a\right)=\frac{1}{4}-\left(\sqrt{a}-\frac{1}{2}\right)^2\le\frac{1}{4}\)

Dấu \(=\)xảy ra khi \(\sqrt{a}-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{4}\). Vậy GTLN của \(A\) là \(\frac{1}{4}\).

\(A=\frac{a^2-b^2}{a-b}=\frac{\left(a-b\right)\left(a+b\right)}{a-b}=a+b\)

\(\Rightarrow A=a+b=a+\frac{1}{a}\ge2\)mà chú ý a>b nên \(1=ab< a^2\Rightarrow a>1\)

Nên dấu bằng của bất đằng thức Cauchy không xảy ra

hay A không có giá trị nhỏ nhất

Ta có: 3¹⁵⁰ = (3²)⁷⁵ = 9⁷⁵

           2²²⁵ = (2³)⁷⁵ = 8⁷⁵

Vì 9⁷⁵ > 8⁷⁵ nên 3¹⁵⁰ > 2²²⁵

Vậy 3¹⁵⁰ > 2²²⁵.

\(M=\frac{a+1}{\sqrt{a}}+\frac{a\sqrt{a}-1}{a-\sqrt{a}}+\frac{a^2-a\sqrt{a}+\sqrt{a}-1}{\sqrt{a}-a\sqrt{a}}\)

\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}{\sqrt{a}.\left(\sqrt{a}-1\right)}+\frac{\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}.\left(1-\sqrt{a}\right)\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow M=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\ge\frac{4\sqrt{a}}{\sqrt{a}}=4\)( BDT Cauchy) mà x khác 1 nên dấu bằng không xảy ra hay M>4

b.\(N=\frac{6}{M}< \frac{6}{4}\)nên N nguyên thì N=0 hoặc N=1

TH1: \(N=0\Leftrightarrow\frac{6}{M}=0\) vô lí

TH2:\(N=1\Leftrightarrow \dfrac{6}{M}=1\Leftrightarrow M=6\)\(\Leftrightarrow\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}=6\Leftrightarrow\left(\sqrt{a}+1\right)^2=6\sqrt{a}\)

\(\Leftrightarrow a-4\sqrt{a}+1=0\Rightarrow\sqrt{a}=2\pm\sqrt{3}\)

Hay \(a=7\pm4\sqrt{3}\)

a)xet tg abc ab^2+ac^2=Bc^2

thay ab=6cm(gt),ac=8cm(gt) ta co

6^2+8^2=BC^2

36+64=BC^2

100=BC^2

10^2=bc^2 suy ra bc=10cm

\(A=\frac{x^2+2x+3}{x^2+4x+4}\Rightarrow A\left(x^2+4x+4\right)=x^2+2x+3\Leftrightarrow x^2\left(A-1\right)+\left(4A-2\right)x+4A-3=0\)(*)

\(\Delta'=\left(2A-1\right)^2-\left(4A-3\right)\left(A-1\right)=3A-2\)

Để phương trình (*) có nghiệm \(x\)thì \(\Delta'\ge0\Rightarrow3A-2\ge0\Leftrightarrow A\ge\frac{2}{3}\).

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