2b)

xét vế trái ta có

=\(\left(\sqrt{x}-\sqrt{y}\right).\dfrac{\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\) \(\left(\sqrt{x}-\sqrt{y}\right).\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)=x-y

3b)

để A<0 \(\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Rightarrow\sqrt{x}-1< 0\)\(\Rightarrow\sqrt{x}< 1\)\(\Rightarrow x< 1\)

a: Ta có: \(\sqrt{9x^2-6x+1}=5\)

\(\Leftrightarrow\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

ngu

b: \(x^2-\dfrac{16}{25}=0\)

\(\Leftrightarrow x^2=\dfrac{16}{25}\)

hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

b) \(x^2-\dfrac{16}{25}=0\Rightarrow x^2=\dfrac{16}{25}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

c) \(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)

\(\Rightarrow\left|\dfrac{1}{2}-x\right|=-\dfrac{28}{5}\)(vô lý do \(\left|\dfrac{1}{2}-x\right|\ge0\))

Vậy \(S=\varnothing\)

b,x²=0+16/25

TH1:x=4/5

TH2:x=-4/5

c,|1/2-x|=-28/5(vì 2/5-6)

bỏ(vì vô lí)

b: BCNN(21;55)=1155

a: Ta có: \(\widehat{xBC}=\widehat{ABM}\)

mà \(\widehat{ABM}=\widehat{NMB}\)

nên \(\widehat{xBC}=\widehat{BMN}\)

b: \(=\dfrac{x^2-x+1-3+1-x^2}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\dfrac{-x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-1}{x^2-x+1}\)

a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

\(M=\left(\dfrac{x+3}{x-3}-\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-3}{x+3}\right):\dfrac{x+3-x-1}{x+3}\)

\(=\dfrac{x^2+6x+9-18+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{2}\)

\(=\dfrac{2x^2}{x-3}\cdot\dfrac{1}{2}=\dfrac{x^2}{x-3}\)

b: Để M nguyên thì \(x^2-9+9⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;3;-3;9;-9\right\}\)

hay \(x\in\left\{4;2;6;0;12;-6\right\}\)

\(b,\widehat{EHG}=360-\widehat{EFG}-\widehat{FGH}-\widehat{FEH}=360-90-60-70=140\\ \Rightarrow x=180-\widehat{EHG}=40\)

Câu c ko thấy cái góc chỗ x tên gì nha

a: \(x=360^0-115^0-70^0-75^0=100^0\)

b: \(x=180^0-\left(360^0-60^0-90^0-70^0\right)=40^0\)