\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

\(b)3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow101x-303=0\)

\(\Leftrightarrow101\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

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!

\(a;\)

\(=0-1\)

\(=-1\)

\(b;\)

\(=0-4\)

\(=-4\)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{15}\)

\(\frac{181\left(x+1\right)}{660}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{181\left(x+1\right)}{660}=\frac{17\left(x+1\right)}{52}\)

\(2353\left(x+1\right)=2805\left(x+1\right)\)

\(2353x+2353=2805x+2805\)

\(2353=2805x+2805-2353x\)

\(2353=452x+2805\)

\(2353-2805=452x\)

\(-452=452x\)

\(x=-1\)

Giải:

Ta có:

\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)

\(\Leftrightarrow\frac{x}{15}+\frac{1}{15}+\frac{x}{7}+\frac{2}{7}+\frac{x}{4}+\frac{4}{4}+6=0\)

\(\Leftrightarrow\frac{x}{15}+\frac{x}{7}+\frac{x}{4}=-\frac{772}{105}\)

\(\Leftrightarrow x\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}\right)=-\frac{772}{105}\)

\(\Leftrightarrow x=-16\)

Vậy phương trình trên có nghiệm là x = -16.

b. Cách làm tương tự.

Chúc bạn học tốt@@

\(x\ne\left\{-10;0\right\}\)

\(\Leftrightarrow200\left(x+10\right)-200x=x\left(x+10\right)\)

\(\Leftrightarrow x^2+10x-2000=0\)

\(\Rightarrow\left[{}\begin{matrix}x=40\\x=-50\end{matrix}\right.\)

Với mọi n nguyên dương ta có:

\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)

Với k nguyên dương thì 

\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)

\(=\sqrt{k+1}-\sqrt{k-1}\)(*)

Đặt A = vế trái. Áp dụng (*) ta có:

\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)

\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)

...

\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)

Cộng tất cả lại

\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)

3. 

Theo bất đẳng thức cô si ta có: 

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

Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)

vế phải đâu?

ko có vế pk

Ủa ko có vế phải thì mình làm bằng niềm tin à? :D

\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)

=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

=> \(x+205=0\)

=> \(x=-205\)

\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)

\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

\(=>x+205=0\)

\(=>x=-205\)

Ta có

\(S=3.\left(\frac{1}{1}.4\right)+3.\left(\frac{1}{4}.7\right)+...+3.\left(\frac{1}{197}.200\right)\)
\(S=3.\left(\frac{1}{1}.4+\frac{1}{4}.7+\frac{1}{7}.10+...+\frac{1}{197}.200\right)\)