a: Xet ΔABC có AD/AB=AE/AC

nên DE//CB

b: DE//BC

AH vuông góc BC

=>AH vuông góc DE

ΔADE cân tại A

mà AH là đường cao

nên AH là trung trực của DE

c: ΔCBA đều

mà BF là trung tuyến

nên BF vuông góc AC

b,\(\left|x\right|=\dfrac{1}{3}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Thay \(x=\dfrac{1}{3},y=-3\) vào B ta có:

\(B=2x^2-3xy+4y^2=2.\left(\dfrac{1}{3}\right)^2-3.\dfrac{1}{3}.\left(-3\right)+4.\left(-3\right)^2=2.\dfrac{1}{9}+3+4.9=\dfrac{2}{9}+3+36=\dfrac{353}{9}\)

Thay \(x=\dfrac{-1}{3},y=-3\) vào B ta có:

\(B=2x^2-3xy+4y^2=2.\left(\dfrac{-1}{3}\right)^2-3.\dfrac{-1}{3}.\left(-3\right)+4.\left(-3\right)^2=2.\dfrac{1}{9}-3+4.9=\dfrac{2}{9}-3+36=\dfrac{299}{9}\)

d,\(D=5\left(x-3\right)+4\left(2-x\right)=5x-15+8-4x=x-7\)

\(x^2=4\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

Thay x=-2 vào D ta có:

\(D=x-7=-2-7=-9\)

Thay x=2 vào D ta có:

\(D=x-7=2-7=-5\)

e,Thay x=2 vào E ta có:

\(\dfrac{2x^2-5x+6}{x-3}=\dfrac{2.2^2-5.2+6}{2-3}=\dfrac{8-10+6}{-1}=-4\)

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

\(F=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{120}{15}=8\)

b: Ta có: \(\left(x-4\right)^2-x\left(x+1\right)\)

\(=x^2-8x+16-x^2-x\)

=-9x+16

a) A=x²+2xy-3x³+2y³+3x³-y³

      = x²+2xy+y³

thay x=5, y=4 ta có:

A=5²+2.5.4+4³

    =129

b) B= xy-x²y²+x⁴y⁴-x⁶y⁶+x⁸y⁸

thay x=-1, y=-1

B=(-1)(-1)-(-1)².(-1)²+(-1)⁴.(-1)⁴-(-1)⁶.(-1)⁶+(-1)⁸.(-1)⁸

    =1-1+1-1+1

    = 1

mik trả lời nhầm, mình xin lỗi

Bài 8:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1--->0,15------>0,05---------->0,15

a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

b. \(V'=0,15.22,4=3,36\left(l\right)\)

c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)

Bài 9:

\(n_{Fe}=\dfrac{2,8}{56}=0,2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2-->0,4---->0,2------>0,2

a. \(V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

b.\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. \(CM_{FeCl_2}=\dfrac{0,2}{0,4}=0,5M\)

Bài 10:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1<--0,2<----0,1<----0,1

a. \(m=m_{Fe}=0,1.56=5,6\left(g\right)\)

b. \(V_{HCl}=\dfrac{0,2}{0,1}=2\left(l\right)=200\left(ml\right)\)

c.\(CM_{FeCl_2}=\dfrac{0,1}{2}=0,05M\)

câu 35) B

câu 15) 

D

câu 18) A