\(a)x+2x+3x+4x+...+100x=-213 b)\frac12x-\frac13=\frac14x-\frac16 c)3.(x-2)+2.(x-1)=10 d)\frac{\mathrm x+1}{\mathrm 3}=\frac{\mathrm x-2}{\mathrm 4} \)
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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) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)
\(A=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left[\frac{\left(x+2\right)^2}{4-x^2}+\frac{4x^2}{4-x^2}-\frac{\left(2-x\right)^2}{4-x^2}\right]:\left[\frac{x\left(x-3\right)}{x^2.\left(2-x\right)}\right]\)
\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{4-x^2}\right]:\left[\frac{x-3}{x\left(2-x\right)}\right]\)
\(A=\frac{4x^2+8x}{4-x^2}:\frac{x-3}{x\left(2-x\right)}\)
\(A=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x^2}{x-3}\)
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) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)
\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)
\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)
\(B=\sqrt{3}\)
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, 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 \(\left\{{}\begin{matrix}\Delta>0\left(luondung\right)\\S\in Z\\P\in Z\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}2m-1\in Z\\-5\in Z\left(tm\right)\end{matrix}\right.\)
\(u=2x\Rightarrow du=2dx\Rightarrow d\left(2x\right)=2dx\Leftrightarrow dx=\dfrac{1}{2}d\left(2x\right)\)
\(\Rightarrow\int f\left(2x\right)dx=\dfrac{1}{2}\int f\left(2x\right).d\left(2x\right)=\dfrac{1}{2}.\left(2.2x.e^{2.2x+1}\right)+C=2x.e^{4x+1}+C\)
Đề là:
\(x^{n+3}y^4:x^7y^n\) hay \(x^{n+3}y^4:\left(x^7y^n\right)\)vậy bạn?