Bài 2:

a: \(2\left(x-4\right)-x+3=0\)

\(\Leftrightarrow2x-8-x+3=0\)

hay x=5

b: \(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

\(a,\) Áp dụng t/c dtsbn:

\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}=\dfrac{5x}{50}=\dfrac{2z}{42}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=12\\z=42\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20};\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{124}{62}=2\\ \Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)

\(c,\) Áp dụng t/c dtsbn

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\\ \Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)

\(d,\) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(xy=54\Rightarrow2k\cdot3k=54\Rightarrow k^2=9\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=9\\x=-6;y=-9\end{matrix}\right.\)

\(e,\) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\Rightarrow x=5k;y=3k\)

\(x^2-y^2=4\Rightarrow25k^2-9k^2=4\Rightarrow16k^2=4\Rightarrow k^2=\dfrac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2};y=\dfrac{3}{2}\\x=-\dfrac{5}{2};y=-\dfrac{3}{2}\end{matrix}\right.\)

\(f,\) Áp dụng t/c dtsbn:

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

\(\Rightarrow\left\{{}\begin{matrix}2x=y+z+1\\2y=x+z+1\\2z=x+y-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y+z=3x-1\\x+y+z=3y-1\\x+y+z=3z+2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3x-1=\dfrac{1}{2}\\3y-1=\dfrac{1}{2}\\3z+2=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=-\dfrac{1}{2}\end{matrix}\right.\)

2:

a: Xét tứ giác DIHK có 

\(\widehat{DIH}=\widehat{DKH}=\widehat{IDK}=90^0\)

Do đó: DIHK là hình chữ nhật

Suy ra: DH=KI(1)

Xét ΔDEF vuông tại D có DH là đường cao ứng với cạnh huyền EF

nên \(DH^2=HE\cdot HF\left(2\right)\)

Từ (1) và (2) suy ra \(IK^2=HE\cdot HF\)

IX
2. j
3. i
4. f
5. c
6. a
7. h
8. e
9. g
10. d

XI
2. part => parts
3. a => an
4. a => an
5. a => the
6. are => will be (không chắc lắm)
7. taking => take
8. are => is

C. 
Bài 1
1. C
2. B
3. C
4. B

(Nên double-check trước khi chép)
 

1.B

2.A

3.B

4.C

5.B

6.D

7.C

8.D

9.B

10.A

e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{-4x^2+x^2-2x+1-x^2-2x-1}{\left(1-x\right)\left(1+x\right)}=\dfrac{-4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}=\dfrac{4x}{x-1}\\ C=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x}{2-x}\\ E=\dfrac{x^2-9-x^2+4x-4-x^2+9}{\left(x-2\right)\left(x+3\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{2-x}{x+3}\)