126 x ( a - a ) x a x 26

126 x 0 x a x 26

0

126 x (a x 1 - a : 1) x a x 26

= 126 x 0 x a x 26

= 0

126 x ( a x ( 1-1) x a x 26

= 126 x a x 0 x a x 26

= 0 x 26 x a

= 0 x a

= 0

126.(a-a).26a=0

126 * { a * 1 - a : 1 } * a 26

=126 * 0 * a * 26

= 0

k nha

2) \(\hept{\begin{cases}^{x^2-xy=y^2-yz}\left(1\right)\\^{y^2-yz=z^2-zx}\left(2\right)\\^{z^2-zx=x^2-xy}\left(3\right)\end{cases}}\)

lấy (2) - (1) suy ra\(2yz=2y^2+xy+xz-x^2-z^2\)

lấy (3) - (1) suy ra \(2xy=zx+yz-z^2+2x^2-y^2\) 

lấy (3) - (2) suy ra \(2zx=xy+yz+2z^2-x^2-y^2\)

cộng lại đc \(yz+xz+xy=0\) do đó \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+xz+xy}{xyz}=0\)

1) \(a=x^2-xy=x\left(x-y\right)\ne0\left(x\ne0,x\ne y\right)\)

mik cần c3 , ai làm giúp mik đc ko 

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

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

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

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

a x b = b x a

( a x b ) x c = a x ( b x c )

a x 1 = 1 x a = a

a x ( b + c ) = a x b + a xc

a : 1 = a

Câu ...a = a sai đề

a : a = 1 

0 : a = 0 

TK mk nha o0o công chúa sinh đôi o0o

Tk mk nha các bn

Ai tk mk.mk tk lại

Bài 1:

a x b = b x a

(a x b) x c = a x (b x c)

a x 1 = 1 x a = a

a x (b + c) = a x b + a x c

a : 1 = a

... : a = a là sai. Nên sửa là: ... : a = 1

a : a = 1

0 : a = 0

axb=bxa

(axb)xc=ax(bxc)

ax1=1xa=a

ax(b+c)=axb+axc

a:1=a

a.a:a=a

0:a=0

Giúp tớ với

\(A=\frac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x+16}{\sqrt{x}+3}\)

+ \(A=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\) \(=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{25}{\sqrt{x}+3}}-6=10-6=4\)

Dấu "=" \(\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}+3=5\Leftrightarrow x=4\)

Vậy \(A=\frac{x+16}{\sqrt{x}+3}\)

Min A = 4 \(\Leftrightarrow x=4\)