a) \(x+2x+3x+...+100x=-213\)

\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)

\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)

\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)

\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)

                                                                    \(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(\Rightarrow3x-6+2x-2=10\)

\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)

a) \(x+2x+3x+4x+...+100x=-213\)

\(x.\left(1+2+3+4+...+100\right)=-213\)

\(x.5050=-213\)

\(x=-\frac{213}{5050}\)

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)

\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)

\(\frac{1}{2}x=-\frac{43}{12}\)

\(x=\frac{-43}{6}\)

c) \(3\left(x-2\right)+2\left(x-1\right)=10\)

\(3x-6+2x-2=10\)

\(3x+2x-\left(6+2\right)=10\)

\(5x-8=10\)

\(5x=18\)

\(x=\frac{18}{5}\)

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

\(\Rightarrow4.\left(x+1\right)=3\left(x-2\right)\)

\(4x+4=3x-6\)

\(\Rightarrow4x-3x=-6-4\)

\(x=-10\)

\(A=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(A=\left[\frac{\left(x+2\right)^2}{4-x^2}+\frac{4x^2}{4-x^2}-\frac{\left(2-x\right)^2}{4-x^2}\right]:\left[\frac{x\left(x-3\right)}{x^2.\left(2-x\right)}\right]\)

\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{4-x^2}\right]:\left[\frac{x-3}{x\left(2-x\right)}\right]\)

\(A=\frac{4x^2+8x}{4-x^2}:\frac{x-3}{x\left(2-x\right)}\)

\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)

\(A=\frac{4x^2}{x-3}\)

giúp mình với !!!

cảm ơn bạn !

a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)

\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)

\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)

\(=\left(4-8+25\right)\sqrt{x^2+1}\)

\(=21\sqrt{x^2+1}\)

b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)

\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)

\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)

\(B=\sqrt{3}\)

Dạ đậy ạ,mong dc gp

a, Ta có : \(x=81\Rightarrow\sqrt{x}=9\)

Thay \(\sqrt{x}=9\)vào biểu thức A ta được : 

\(A=\frac{2}{9+1}=\frac{2}{10}=\frac{1}{5}\)

b, Ta có : \(P=\frac{B}{A}\)hay\(P=\frac{\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}}{\frac{2}{\sqrt{x}+1}}\)

\(=\frac{1+\sqrt{x}}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c, Ta có \(\frac{1}{2}=\frac{\sqrt{x}}{2\sqrt{x}}\)mà \(\sqrt{x}< \sqrt{x}+1\)

nên \(P>\frac{1}{2}\)

a) \(A=\frac{2}{\sqrt{x}+1}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}\)

b) \(B=\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}\)

\(=\frac{1+\sqrt{x}}{\left(1+\sqrt{x}\right)\sqrt{x}}=\frac{1}{\sqrt{x}}\)

\(\Rightarrow P=\frac{B}{A}=\frac{1}{\sqrt{x}}\div\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c) Ta có: \(P=\frac{\sqrt{x}+1}{2\sqrt{x}}=\frac{1}{2}+\frac{1}{\sqrt{x}}+\frac{1}{2}+0=\frac{1}{2}\)

=> P>1/2

a)Thay $x=81(\mathrm{t}\mathrm{m}dk)$ vào biều thức $A$, ta được \mathrm{A}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}A=81​+12​=9+12​=51​
Vậy $x=81$ thì \mathrm{A}=\frac{1}{5}A=51​

b) B=\frac{1}{\sqrt{x}(\sqrt{x}+1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)1​+x​(x​+1)x​​
B=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)x​+1​
B=\frac{1}{\sqrt{x}}B=x​1​
P=\frac{1}{\sqrt{x}} \div \frac{2}{\sqrt{x}+1}P=x​1​÷x​+12​
P=\frac{1}{\sqrt{x}} \cdot \frac{\sqrt{x}+1}{2}P=x​1​⋅2x​+1​
P=\frac{\sqrt{x}+1}{2 \sqrt{x}}P=2x​x​+1​

c) Ta có P-\frac{1}{2}=\frac{\sqrt{x}+1}{2 \sqrt{x}}-\frac{1}{2}=\ldots=\frac{1}{2 \sqrt{x}}P−21​=2x​x​+1​−21​=…=2x​1​
Ta có $x>0$ nên \frac{1}{2 \sqrt{x}}>0 \Rightarrow P-\frac{1}{2}>0 \Rightarrow P>\frac{1}{2}2x​1​>0⇒P−21​>0⇒P>21​
 

a, Để pt trên có 2 nghiệm pb thì \(\Delta>0\)

\(\Delta=4m^2-4m+1+20=\left(2m-1\right)^2+20>0\forall m\)( đpcm )

Câu a:  Ta có \(\Delta\)= (1-2m)²-4.1.5= (2m-1)²+20>0 với mọi m

    ⇒Phương trình luôn có 2 nghiệm phân biệt với mọi m

Câu b:

Để phương trình có 2 nghiệm nguyên thì  \(\left\{{}\begin{matrix}\Delta>0\left(luondung\right)\\S\in Z\\P\in Z\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}2m-1\in Z\\-5\in Z\left(tm\right)\end{matrix}\right.\)  

\(u=2x\Rightarrow du=2dx\Rightarrow d\left(2x\right)=2dx\Leftrightarrow dx=\dfrac{1}{2}d\left(2x\right)\)

\(\Rightarrow\int f\left(2x\right)dx=\dfrac{1}{2}\int f\left(2x\right).d\left(2x\right)=\dfrac{1}{2}.\left(2.2x.e^{2.2x+1}\right)+C=2x.e^{4x+1}+C\)

a)\(\left\{{}\begin{matrix}2x-4y=10\\2x-y=7\end{matrix}\right.\\ \left\{{}\begin{matrix}-3y=3\\2x-y=7\end{matrix}\right.\\ \left\{{}\begin{matrix}y=-1\\2x-\left(-1\right)=7\end{matrix}\right.\\ \left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\)

b)\(-\dfrac{1}{4}x^2=\dfrac{1}{2}x-2\\ \Leftrightarrow-\dfrac{1}{4}x^2-\dfrac{1}{2}x+2=0\\ \left[{}\begin{matrix}x_1=2\\x_2=-4\end{matrix}\right.\\ \left[{}\begin{matrix}y_1=-1\\y_2=-4\end{matrix}\right.\)

Đề là:

\(x^{n+3}y^4:x^7y^n\) hay \(x^{n+3}y^4:\left(x^7y^n\right)\)vậy bạn?