khiếp nhiều thế

a) \(\frac{x+1}{x-1}-\frac{3x+1}{x^2-x}=\frac{1}{x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 1

<=> \(\frac{x\left(x+1\right)}{x\left(x-1\right)}-\frac{3x+1}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}=0\)

<=> \(\frac{x^2+x}{x\left(x-1\right)}-\frac{3x+1}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}=0\)

<=> \(\frac{x^2+x-3x-1-x+1}{x\left(x-1\right)}=0\)

<==> \(\frac{x^2-3x}{x\left(x-1\right)}=0\)

=> x² - 3x = 0

<=> x( x - 3 ) = 0

<=> x = 0 ( ktm ) hoặc x = 3 ( tm )

Vậy phương trình có nghiệm x = 3

b) \(\frac{5x}{3x-6}-\frac{2x-3}{2x-4}=\frac{1}{2}\)

ĐKXĐ : x ≠ 2

<=> \(\frac{5+x}{3\left(x-2\right)}-\frac{2x-3}{2\left(x-2\right)}-\frac{1}{2}=0\)

<=> \(\frac{2\left(5+x\right)}{6\left(x-2\right)}-\frac{3\left(2x-3\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=0\)

<=> \(\frac{10+2x}{6\left(x-2\right)}-\frac{6x-9}{6\left(x-2\right)}-\frac{3x-6}{6\left(x-2\right)}=0\)

<=> \(\frac{10+2x-6x+9-3x+6}{6\left(x-2\right)}=0\)

<=> \(\frac{-7x+25}{6\left(x-2\right)}=0\)

=> -7x + 25 = 0 <=> x = 25/7 ( tm )

Vậy phương trình có nghiệm x = 25/7

c) \(\frac{6x-x^2}{x^2-2x}+\frac{x}{x-2}=\frac{3}{x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 2

<=> \(\frac{6x-x^2}{x\left(x-2\right)}+\frac{x^2}{x\left(x-2\right)}-\frac{3\left(x-2\right)}{x\left(x-2\right)}=0\)

<=> \(\frac{6x-x^2+x^2-3x+6}{x\left(x-2\right)}=0\)

<=> \(\frac{3x+6}{x\left(x-2\right)}=0\)

=> 3x + 6 = 0 <=> x = -2 ( tm )

Vậy phương trình có nghiệm x = -2

d) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 2

<=> \(\frac{2}{x\left(x-2\right)}-\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=0\)

<=> \(\frac{2}{x\left(x-2\right)}-\frac{x^2+2x}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=0\)

<=> \(\frac{2-x^2-2x-x+2}{x\left(x-2\right)}=0\)

<=> \(\frac{-x^2-3x+4}{x\left(x-2\right)}=0\)

=> -x² - 3x + 4 = 0

<=> x² + 3x - 4 = 0

<=> ( x - 1 )( x + 4 ) = 0

<=> x = 1 hoặc x = -4 ( thỏa mãn )

Vậy tập nghiệm của phương trình là : S = { 1 ; -4 }

e) \(\frac{x+2}{x-2}-\frac{6}{x+2}=\frac{x^2}{x^2-4}\)

ĐKXĐ : x ≠ ±2

<=> \(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{6x-12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2+4x+4-6x+12-x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{-2x+16}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x + 16 = 0 <=> x = 8 ( tm )

Vậy phương trình có nghiệm x = 8

f) \(\frac{3}{1-3x}=\frac{2}{1+3x}-\frac{7-5x}{1-9x^2}\)( chắc là như này )

ĐKXĐ : x ≠ ±1/3

<=> \(\frac{3\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}-\frac{2\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{3+9x}{\left(1-3x\right)\left(1+3x\right)}-\frac{2-6x}{\left(1-3x\right)\left(1+3x\right)}+\frac{7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{3+9x-2+6x+7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{10x+8}{\left(1-3x\right)\left(1+3x\right)}=0\)

=> 10x + 8 = 0 <=> x = -4/5 ( tm )

Vậy phương trình có nghiệm x = -4/5

Đặt \(A=2^{4^n}+4\)

Vì 4ⁿ \(⋮4\)

=> Đặt 4ⁿ = 4k (k \(\inℕ^∗\))

Khi đó A = \(2^{4k}+4=16^k+4=\left(...6\right)+4=\left(...0\right)\)

=> \(A⋮10\)

Ta có : \(x\left(x+4\right)\left(x+2\right)^2=45\)

\(\Rightarrow\left(x^2+4x\right)\left(x^2+4x+4\right)=45\left(1\right)\)

Đặt \(x^2+4x+2=a\)

Thay \(a\)vào \(\left(1\right)\), khi đó :

\(\left(1\right)\Leftrightarrow\left(a-2\right)\left(a+2\right)=45\)

\(\Rightarrow a^2-4=45\)

\(\Rightarrow a^2=49\)

\(\Rightarrow\orbr{\begin{cases}a=7\\a=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+4x+2=7\\x^2+4x+2=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+4x-5=0\\x^2+4x+4=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\left(x+5\right)\left(x-1\right)=0\\\left(x+2\right)^2=-5\end{cases}}\)(Vô lí, do \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}\)

Vậy \(x\in\){\(-5;1\)}

A B C D O

a. ta có \(\frac{AO}{OC}=\frac{AB}{CD}=\frac{1}{2}\Rightarrow\hept{\begin{cases}AO=4cm\\OC=8cm\end{cases}}\) theo pythago ta có \(\hept{\begin{cases}OB=\sqrt{AB^2-AO^2}=3cm\\OD=\sqrt{CD^2-CO^2}=6cm\end{cases}}\Rightarrow BD=9cm\)

b, diện tích ABCD \(=\frac{1}{2}AC.BD=\frac{1}{2}.12.9=54cm^2\)

c. ta có diện tích ABCD \(=54=\frac{1}{2}\left(AB+CD\right).h\Rightarrow h=\frac{36}{7}cm\)

a)Ta có : \(4x^2=1\)

\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)

Thay \(x=\frac{1}{2}\)vào B , ta được:

\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)

Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)

b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)

\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)

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

\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)

\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)

\(=\frac{x}{x+1}\)

Vậy \(M=\frac{x}{x+1}\)

c)Ta có: \(x< x+1\forall x\)

\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)

Vậy với mọi \(x\ne-1\)thì \(M< 1\)

a, Ta có \(a^3+2a^2-3=a^3-a^2+3a^2-3=a^2\left(a-1\right)+3\left(a-1\right)\left(a+1\right)=\left(a-1\right)\left(a^2+3a+3\right)\)

b, Để B có giá trị là số nguyên tố thì hoặc a-1=1 hoặc a^2+3a+3=1

nếu\(a-1=1\Rightarrow a=2\Rightarrow B=13\)thỏa mãn

nếu\(\left(a^2+3a+3\right)=1\Leftrightarrow a^2+3a+2=0\Leftrightarrow\orbr{\begin{cases}a=-1\Rightarrow B=-2\\a=-2\Rightarrow B=-3\end{cases}}\)không thỏa mãn

vậy a=2 thì B là số nguyên tố