Bài 1: 

a: Xét ΔABI và ΔACI có

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

2.

\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)

\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)

\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)

\(\Leftrightarrow cosx\left(cosx+sinx\right)\left(cosx-sinx-1\right)=0\)

\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x=k2\pi\end{matrix}\right.\)

Bài 1: 

Vì (d)//y=-2x+1 nên a=-2

Vậy: y=-2x+b

Thay x=1 và y=2 vào (d),ta được:

b-2=2

hay b=4

Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

gfrưerrrrrrrrrrr

Bài 2

5 C

Bài 3

1 D

6 C

Còn lại ol r nhé

2) 5. C

3) 2. D

6. C

Còn lại ok nha

ko bik 

Bài 2:

Ta có:\(2\sqrt{48}< 2\sqrt{49}\) ; 

         \(3\sqrt{27}>3\sqrt{25}\)

mà \(2\sqrt{49}< 3\sqrt{25}\left(14< 15\right)\)

\(\Rightarrow3\sqrt{27}>3\sqrt{25}>2\sqrt{49}>2\sqrt{48}\)

\(\Rightarrow3\sqrt{27}>2\sqrt{48}\)

b)

Ta có:\(\sqrt{50}+\sqrt{2}>\sqrt{49}+\sqrt{1}\) 

        \(\sqrt{50+2}< \sqrt{64}\)

mà \(\sqrt{49}+\sqrt{1}=\sqrt{64}\left(8=8\right)\)

\(\Rightarrow\sqrt{50}+\sqrt{2}>8>\sqrt{50+2}\)

\(\Rightarrow\sqrt{50}+\sqrt{2}>\sqrt{50+2}\)

Bài 3:

a. \(\sqrt{9\left(x+3\right)}-\dfrac{1}{4}\sqrt{16\left(x+3\right)}+\sqrt{x+3}=6\)

\(\Leftrightarrow3\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+3}=6\)

\(3\sqrt{x+3}=6\)

\(\sqrt{x+3}=2\)

\(\left\{{}\begin{matrix}2>0\left(ld\right)\\x+3=4\end{matrix}\right.\Leftrightarrow x=1\)

b. \(\sqrt{2x-1}=3\)

\(\left\{{}\begin{matrix}3>0\left(ld\right)\\2x-1=9\end{matrix}\right.\Leftrightarrow x=5\)

Bài 2:

\(a,B=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+2\right]:\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-1\right]=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\\ b,B< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-1< 0\\ \Leftrightarrow\dfrac{\sqrt{x}+2-\sqrt{x}+1}{\sqrt{x}-1}< 0\\ \Leftrightarrow\dfrac{3}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Bài 3:

\(a,ĐK:x\ge-3\\ PT\Leftrightarrow3\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+3}=6\\ \Leftrightarrow3\sqrt{x+3}=6\\ \Leftrightarrow\sqrt{x+3}=2\\ \Leftrightarrow x+3=4\\ \Leftrightarrow x=1\left(tm\right)\\ b,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow2x-1=9\\ \Leftrightarrow x=5\left(tm\right)\)

1:

a: =12/10-7/10=5/10=1/2

b: \(=\dfrac{4}{13}-\dfrac{4}{13}+\dfrac{-5}{11}-\dfrac{6}{11}=-\dfrac{11}{11}=-1\)

2: 

a: x+2/7=-11/7

=>x=-11/7-2/7=-13/7

b: (x+3)/4=-7/2

=>x+3=-14

=>x=-17