a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)

\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)

\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)

c: 2x-3căn x-5=0

=>2x-5căn x+2căn x-5=0

=>2căn x-5=0

=>x=25/4

Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)

\(A=\left(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}\right):\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\)

\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)

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

đk : xkhác -1 ; 1 

\(A=\left(\dfrac{1+x-1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{x-x+1}{x\left(x-1\right)}\right)=\dfrac{x}{\left(x+1\right)\left(x-1\right)}:\dfrac{1}{x\left(x-1\right)}=\dfrac{x^2}{x+1}\)

\(A=\dfrac{x-1+1}{\left(x+1\right)\left(x-1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

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

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

\(a.\) Với  \(a+b+c=0\)  thì  \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)

\(b.\)   Công thức tổng quát:  \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

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

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

\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)

\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)

Do đó, suy ra được:  \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)

`đk:x ne +-3,x ne -2`

`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`

`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`

`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`

`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`

`=(3x+6)/((x-3)(x+2))`

`=3/(x-3)`

`b)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\) 

`=>B=3/(2-3)=-3`

`c)B=-3/5`

`<=>3/(x-3)=3/(-5)`

`<=>x-3=-5`

`<=>x=-2(l)`

`d)B<0`

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

Mà `3>0`

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

a) đk: \(x\ne\pm3\)

 \(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)

= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)

= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)

= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)

b) Để \(\left|2x+1\right|=5\)

<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)

Thay x = 2, ta có;

B = \(\dfrac{3}{2-3}=-3\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)

<=> x - 3 = -5

<=> x = -2

d) Để B < 0

<=> \(\dfrac{3}{x-3}< 0\)

<=> x - 3 < 0

<=> x < 3

a)\(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\\ =\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x+3}\)

\(=\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)

b)\(\left|2x+1\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\left(loại\right)\end{matrix}\right.\)

với x=2 gt của B là

\(B=\dfrac{3}{2-3}=-3\)

c)\(B=\dfrac{3}{x-3}=-\dfrac{3}{5}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)

d) \(B=\dfrac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

tự kết luận mỗi câu

a: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}+4}{x\sqrt{x}-3x+2\sqrt{x}}-\dfrac{3\sqrt{x}+3}{-x+\sqrt{x}+2}\right):\left(\dfrac{x-\sqrt{x}-6}{x-3\sqrt{x}}-\dfrac{x-2\sqrt{x}}{x-4\sqrt{x}+4}\right)+\sqrt{x}\)

\(=\left(\dfrac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{3}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)+\sqrt{x}\)

\(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}+\sqrt{x}\)

\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}+\sqrt{x}\)

\(=-\sqrt{x}-1+\sqrt{x}\)

=-1

đkxd: \(x\ne\left\{\pm3\right\}\)

a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)

=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)

=\(\frac{x+12}{x-3}\)

b)|2x+1|=5

<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2

với x=-3 thì B=\(\frac{-3}{2}\)

với x=2 thì B=-14

Phần a,b,c bạn có thể tham khảo bài bên dưới. 

Phần d.

ĐKXĐ: $x\geq 0; x\neq 4$

$A>5\Leftrightarrow \frac{x+9}{2\sqrt{x}}>5$ ($x> 0$)

$\Leftrightarrow x+9> 10\sqrt{x}$

$\Leftrightarrow x-10\sqrt{x}+9>0$

$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-9)>0$

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}-1>0\\ \sqrt{x}-9>0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}-1<0\\ \sqrt{x}-9<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>1\\ x>81\end{matrix}\right.\\ \left\{\begin{matrix} 0\leq x< 1\\ 0\leq x< 81\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x>81\\ 0\leq x< 1\end{matrix}\right.\)

Kết hợp với đkxđ suy ra $x>81$ hoặc $0< x< 1$

a

Với: x \(\ge0,x\) \(\ne4\) có:

\(A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right):\left(\dfrac{\left(\sqrt{x}+2\right)^2}{x-4}-\dfrac{\left(\sqrt{x}-2\right)^2}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4}{x-4}-\dfrac{x-4\sqrt{x}+4}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-6\sqrt{x}}{x-4}\right)\)

\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{2\sqrt{x}}{x-4}\right)\)

\(=\dfrac{\left(x+9\right)\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}=\dfrac{x+9}{2\sqrt{x}}\)

b

Giải \(x^2-5x+4=0\)

Nhẩm nghiệm: a + b + c = 0 (1 - 5 + 4 = 0)

\(\Rightarrow x_1=1;x_2=\dfrac{c}{a}=\dfrac{4}{1}=4\)

Thay x = 1 vào A:

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

Thay x = 4 vào A:

\(A=\dfrac{4+9}{2.\sqrt{4}}=\dfrac{13}{2.2}=\dfrac{13}{4}\)

c

ĐK: x > 0

\(A=0\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}=0\)

=> \(x+9=0\Rightarrow x=-9\) (không thỏa mãn)

Vậy không xác định được giá trị x

d

ĐK: x > 0 

\(A>5\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}>5\)

\(\Leftrightarrow x+9>5.2\sqrt{x}\Leftrightarrow x+9>10\sqrt{x}\)

\(\Leftrightarrow\left(x+9\right)^2>\left(10\sqrt{x}\right)^2=100x\)

<=> \(x^2+18x+81-100x>0\)

<=> \(x^2-82x+81>0\)

<=> \(x^2-81x-x+81>0\)

<=> \(x\left(x-81\right)-\left(x-81\right)>0\)

<=> \(\left(x-1\right)\left(x-81\right)>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1>0\\x-81>0\end{matrix}\right.\\\left[{}\begin{matrix}x-1< 0\\x-81< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x>81\end{matrix}\right.\\\left[{}\begin{matrix}x< 1\\x< 81\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>81\\x< 81\end{matrix}\right.\)

Vậy để A > 5 thì x > 81 và 0 < x < 81

Hầy mình không nghĩ lớp 7 đã phải làm những bài biến đổi như thế này. Cái này phù hợp với lớp 8-9 hơn.

1.

Đặt $x^2-y^2=a; y^2-z^2=b; z^2-x^2=c$. 

Khi đó: $a+b+c=0\Rightarrow a+b=-c$

$\text{VT}=a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$

$=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$

$=3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)$

$=3(x-y)(y-z)(z-x)(x+y)(y+z)(x+z)$

$=3.4(x-y)(y-z)(z-x)=12(x-y)(y-z)(z-x)$

Ta có đpcm.

Bài 2:

Áp dụng kết quả của bài 1:

Mẫu:

$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x-y)(y-z)(z-x)(x+y)(y+z)(z+x)=3(x-y)(y-z)(z-x)(1)$

Tử: 

Đặt $x-y=a; y-z=b; z-x=c$ thì $a+b+c=0$

$(x-y)^3+(y-z)^3+(z-x)^3=a^3+b^3+c^3$

$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=3abc$

$=3(x-y)(y-z)(z-x)(2)$

Từ $(1);(2)$ suy ra \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}=1\)

Bài 3:

\(ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{2^2-2}{2}=1\)

Do đó:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=\frac{1}{abc}\)

Ta có đpcm.