1. a. \(\sqrt{25}=\pm5\)

b. \(\sqrt{0,0001}=\pm0,01\)

c. \(\sqrt{\dfrac{9}{25}}=\pm\dfrac{3}{5}\)

d. \(\sqrt{-6}=\varnothing\)

2. a. \(\sqrt{25-9}=\sqrt{16}=\pm4\)

b. \(\sqrt{0,01}-\sqrt{0,25}=\pm0,1-\pm0,5=\left[{}\begin{matrix}-0,4\\0,4\\0,6\\-0,6\end{matrix}\right.\)

c tương tự nhé

Bài 3:

a) \(\sqrt{6}< \sqrt{16}=4\)

b) \(\sqrt{11.3}=\sqrt{33}< \sqrt{44}\)

Bài 4:

a) \(x^2+16=25\)

\(\Rightarrow x^2=25-16=9\Rightarrow x=\pm3\)

b) \(\left|\sqrt{x}-3\right|+3=9\left(đk:x\ge0\right)\)

\(\Rightarrow\left|\sqrt{x}-3\right|=9-3=6\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-3=6\\\sqrt{x}-3=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=9\\\sqrt{x}=-3\left(VLý\right)\end{matrix}\right.\)\(\Rightarrow x=81\left(tm\right)\)

\(A=\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)

\(=\left(4x-1\right)\left(3x+1\right)-\left(5x+x-4\right)\left(x-3\right)\)

\(=12x^2+4x-3x-1-6x^2+4x+18x-12\)

\(=18x^2+19x-13\)

\(a,y=kx\Leftrightarrow8=6k\Leftrightarrow k=\dfrac{8}{6}=\dfrac{4}{3}\\ b,y=\dfrac{4}{3}x\\ c,x=12\Leftrightarrow y=\dfrac{4}{3}\cdot12=16\)

f(-6)=17

50->f(-1)=2

Lời giải:

Kẻ tia $Bt\parallel Ax(1)$ (như hình vẽ)

Ta có:

$\widehat{ABt}=\widehat{A}$ (2 góc so le trong)

$\Rightarrow \widehat{ABC}-\widehat{ABt}=\widehat{ABC}-\widehat{A}$

Hay $\widehat{tBC}=\widehat{A}+\widehat{C}-\widehat{A}=\widehat{C}$

Mà 2 góc này ở vị trí so le trong nên $Bt\parallel Cy(2)$

Từ $(1); (2)\Rightarrow Ax\parallel Cy$ (đpcm)

Hình vẽ: