Thích ôm t đúng k trùi :>>

Bài 2

\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2}=7\Leftrightarrow\left|2x-1\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Bài 1:

1) \(A=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}=2+\sqrt{3}-\sqrt{3}-1=1\)

2) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}-3\sqrt{x-2}=-1\Leftrightarrow-2\sqrt{x-2}=-1\Leftrightarrow x-2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)

Bài 2:

a) \(A=\dfrac{2\sqrt{25}-1}{\sqrt{25}-1}=\dfrac{2.5-1}{5-1}=\dfrac{9}{4}\)

b) \(B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

c) \(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}< 1\)

\(\Leftrightarrow\dfrac{3}{\sqrt{x}+1}>1\Leftrightarrow0< \sqrt{x}+1< 3\Leftrightarrow0\le x\le4,x\ne1\)

Bài 1:

1) \(=\dfrac{2}{1}+\sqrt{3}+\sqrt{3}=2+2\sqrt{3}\)

2) Cần đặt chân thang cách tường: \(3,5.cos75^0\approx0,9\left(m\right)\)

Bài 2:

a)\(A=\dfrac{x+5\sqrt{x}}{x-25}=0\Leftrightarrow x+5\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}+5\right)=0\Leftrightarrow x=0\left(do.\sqrt{x}+5\ge5>0\right)\)

b) \(B=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

c) \(P=\dfrac{\sqrt{x}}{\sqrt{x}+3}.\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=1-\dfrac{8}{\sqrt{x}+3}< 1\)

\(25cm=0,25m;7cm=0,07m\)

\(\left\{{}\begin{matrix}p_1=d.h_1=10000.0,25=2500\left(Pa\right)\\p_2=d.h_2=10000.\left(0,25-0,07\right)=1800\left(Pa\right)\end{matrix}\right.\)

\(\left(x-y\right)^2=16\Rightarrow x^2-2xy+y^2=16\Rightarrow x^2+y^2=16+2xy\)

\(x-y=4\Rightarrow\left(x-y\right)^2=16\Rightarrow P=x^2+y^2=16+2xy=16+2.5=26\)

Chọn B

\(1\left(\dfrac{km}{ph}\right)=60\left(\dfrac{km}{h}\right),30ph=\dfrac{1}{2}h\)

\(S=t_1.v_1=t_2.v_2\Rightarrow\left(t_2+\dfrac{1}{2}\right).50=t_2.60\Rightarrow t_2=2,5h\)

Nơi 2 xe gặp nhau cách B: \(180-t_2.v_2=180-2,5.60=30\left(km\right)\)

a) Xét ΔABD và ΔAKD có:

AB=AK(gt)

\(\widehat{BAD}=\widehat{KAD}\)(AD là phân giác)

AD chung

=> ΔABD=ΔAKD(c.g.c)

b) Ta có: \(M\in AB,AM=AC,AC>AB\)

=> B nằm giữa A và M

\(\Rightarrow AB+BM=AM\)

Mà \(AK+KC=AC\) và \(AK=AB,AC=AM\)

\(\Rightarrow BM=KC\)