Ta có:

\(\frac{S_{BDM}}{S_{BDC}}=\frac{BM}{BC}=\frac{1}{3}\left(1\right)\)

Ta lại có

\(\hept{\begin{cases}\frac{S_{AIB}}{S_{BIM}}=\frac{AI}{MI}=\frac{1}{2}\\\frac{S_{ADI}}{S_{MDI}}=\frac{AI}{MI}=\frac{1}{2}\end{cases}}\)

\(\Rightarrow S_{BDM}=S_{BIM}+S_{DIM}=2S_{AIB}+2S_{ADI}=2S_{ABD}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{2S_{ABD}}{S_{BDC}}=\frac{1}{3}\)

\(\Rightarrow\frac{S_{ABD}}{S_{BDC}}=\frac{1}{6}=\frac{AD}{DC}\)

\(\Rightarrow\frac{AD}{AC}=\frac{1}{7}\)

a) Áp dụng ĐL đường phân giác trong tam giác, ta có: 

\(\frac{AM}{HM}=\frac{AC}{HC}\); \(\frac{BN}{HN}=\frac{AB}{AH}\).

Dễ thấy \(\Delta\)AHB ~ \(\Delta\)CHA (g.g): \(\frac{AC}{AB}=\frac{HC}{AH}\Rightarrow\frac{AC}{HC}=\frac{AB}{AH}\)

Do đó: \(\frac{AM}{HM}=\frac{BN}{HN}\)=> MN // AB (ĐL Thales đảo) (đpcm).

b) Áp dụng hệ quả ĐL Thales: \(\frac{MO}{MI}=\frac{AO}{AN}\)(Do NI//AM); \(\frac{MO}{MB}=\frac{NO}{AN}\)

\(\Rightarrow\frac{MO}{MI}+\frac{MO}{MB}=\frac{AO+NO}{AN}=\frac{AN}{AN}=1\Leftrightarrow\frac{1}{MI}+\frac{1}{MB}=\frac{1}{MO}\)(đpcm). 

Câu 1: Rút gọn

a. (x+y)²  + (x-y)²

=x²+2xy+y²+x²-2xy+y²=2x²+2y²

b. 2.(x-y) . (x+y) + (x+y)² + (x-y)²

=2.(x²-y²)+2x²+2y²=4x²

c. (x-y+z)² + (z-y)² +2.(x-y+z) . (z-y)

=x²+y²+z²-2xy-2yz+2zx+z²-2yz+y²+2.(xz-xy-yz+y²+z²-zy)

=x²+2y²+2z²-2xy+2zx-4yz+2xz-2xy-4yz+2y²+2z²

=x²+4y²+4z²-4xy-8yz+4xz

Câu 2: Chứng minh

(ac+bd)² + (ad-bc)²=a²c²+2abcd+b²d²+a²d²-2abcd+b²c²= a²c²+b²d²+a²d²+b²c² =(a²+b²) . (c²+d²) 

Câu 1: 

a. \(\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=2\left(x^2+y^2\right)\)

b. \(2\left(x-y\right)\left(x+y\right)+\left(x+y^2\right)+\left(x-y\right)^2\)

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

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

\(=\left(2x\right)^2\)

\(=4x^2\)

 \(A=100^2-99^2+98^2-97^2+.......+2^2-1^2\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+.......+\left(2^2-1^2\right)\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+.......+\left(2-1\right)\left(2+1\right)\)

\(=1\left(100+99\right)+1\left(98+97\right)+.......+1\left(2+1\right)\)

\(=3+7+.......+195+199\)

Số số hạng là : 

            199 - 3 : 4 + 1 = 50(số)

Tổng A là : 

             (199 + 3) x 50 : 2 = 5050 

\(B=3\left(2^2+1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1\)

\(=\left(4-1\right)\left(2^2+1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^8+1\right).......\left(2^{64}+1\right)+1\)

\(...........................\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1=2^{128}\)

Xét thí nghiệm 1:

\(PTHH:Mg+2HC1->FeCI_2+H_2\)               (1)

Giả sử Fe phản ứng hết -> Chất rắn là \(FeCI_2\)

\(\Rightarrow n_{Fc}=n_{FeCI_2}=n_{h_2}=\frac{3,1}{127}\approx0,024\left(mol\right)\)

Xét thí nghiệm 2:

\(PTHH:Mg+2HCI->MgCI_2+H_2\)(2)

         \(Fe+2HCI->FeCI_2+H_2\)          (3)

Ta thấy :Ngoài a gam Fe như thí nghiệm 1 cộng với b gam Mg mà chỉ giải phóng :

\(n_{H_2}=\frac{0,0448}{22,4}=0,024\left(mol\right)\)

-> Chứng tỏ TH1:Fe dư HCI hết :

Ta có \(n_{HCI}\left(TN1\right)=n_{HCI}\left(TN2\right)=2_{n_{H2}}=2.0,02=0,04\left(mol\right)\)

TH1:

\(n_{Fe\left(pư\right)}=n_{nFeCI_2}=\frac{1}{2}n_{HCI}=\frac{1}{2}.0,04=0,02\left(mol\right)\)

\(\Rightarrow m_{fe\left(dư\right)}=3,1-0,02.127=0,56\left(gam\right)\)

     \(m_{Fe\left(dư\right)}=0,02.56=1,12\left(gam\right)\)

\(\Rightarrow m_{Fe}=a=0,56+1,12=1,68\left(gam\right)\)

TN2:

Áp dụng ĐLBTKL :

\(a+b=3,34+0,02.2-0,04.36,5=1,92\left(g\right)\)

Mà \(a=1,68gam->b=1,92-1,68=0,24\left(g\right)\)

P/s:Thằng lười :v

ủa sao thí nghiệm 1 lại có mg vậy, vô lý quá

Height line = chiều cao giữa các dòng

:))

nghĩa toán học cơ bạn

\(\Rightarrow x^2+y^2-3xxy=0\)

\(\Rightarrow x^2-2xy+y^2-xy=0\)

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

\(\Rightarrow x-y=\sqrt{xy}=\sqrt{x}.\sqrt{y}\)

\(\Rightarrow x=\sqrt{x}.\sqrt{y}+y=\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)

\(\Rightarrow y=x-\sqrt{x}.\sqrt{y}=\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\)

\(\Rightarrow\frac{x}{y}=\frac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}\)