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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

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

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

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`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

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\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}=\frac{a+b-c+a+c-b+b+c-a}{c+b+a}=\frac{a+b+c}{a+b+c}=1\)

=>a+b-c=c=>a+b=2c

   a+c-b=b=>a+c=2b

   b+c-a=a=>b+c=2a

Lại có: \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

=>\(A=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}\)

=>\(A=\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}=\frac{2c.2a.2b}{b.c.a}=\frac{8.\left(a.b.c\right)}{a.b.c}=8\)

Vậy A=8

1)Ta có:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)(đpcm)

Ta có:A=\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)

\(\Rightarrow A=\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{a+c}=\frac{a+c+b}{b+c+a+b+a+c}\)\(\Rightarrow A=\frac{a+b+c}{2a+2b+2c}=\frac{\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy A=\(\frac{1}{2}\)

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)=k

\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

Mà x-2y+3z=-10

Hay (2k+1)-2\(\cdot\)(3k+2)+3\(\cdot\)(4k+3)=-10

2k+1-6k-4+12k+9=-10

8k+6=-10

8k=(-10)-6

8k=-16

k=(-16):8

k=(-2)

\(\Rightarrow x=\left(-2\right)\cdot2+1=-3\)

\(y=\left(-2\right)\cdot3+2=-4\)

\(z=\left(-2\right)\cdot4+3=-5\)

a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)

\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)

\(=\frac{45}{4}-\frac{223}{28}\)

\(=\frac{315}{28}-\frac{223}{28}\)

\(=\frac{23}{7}\)

b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)

   \(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)

   \(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)

   \(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)

    \(=\frac{69}{8}\)

1a) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)

\(=11\frac{1}{4}-2\frac{5}{7}-5\frac{1}{4}\)

\(=11\frac{1}{4}-5\frac{1}{4}-2\frac{5}{7}\)

\(=6-2\frac{5}{7}=3\frac{2}{7}=\frac{23}{7}\)

2a) \(x\div8,25=\left(-2,2\right)\)

\(\Leftrightarrow x=-18,15\)

\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)\)

Mà a+b+c = 0 nên a + c = -b

                             a + b = -c

                             b + c = -a

\(A=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)

thanks bạn nhiều nha

2) \(VT=\left(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\right)+3\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

Xét \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)

Áp dụng bất đẳng thức Cauchy dạng phân thức 

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\) (1) 

Xét \(3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

Áp dụng bất đẳng thức Cauchy dạng phân thức 

\(\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)

\(\Rightarrow3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge3.\frac{3}{2}=\frac{9}{2}\) (2) 

Từ (1) và (2) 

\(\Rightarrow VT\ge\frac{9}{2}+\frac{3}{2}=6\) ( đpcm ) 

Dấu " = " xảy ra khi \(a=b=c=1\)

cám ơn nhiều.

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