a) Diện tích tam giác ABC (Heron)

\(S_{ABC}=\frac{1}{4}\sqrt{\left(AB+BC+AC\right)\left(AB+BC-AC\right)\left(BC+AC-AB\right)\left(AC+AB-BC\right)}\)

\(S_{ABC}=\frac{1}{4}\sqrt{\left(6+10+8\right)\left(6+10-8\right)\left(10+8-6\right)\left(8+6-10\right)}=24\left(cm^2\right)\)

b)Xét tam giác ABC có 

\(BC^2=10^2=100\left(cm\right)\)

\(AB^2+AC^2=6^2+8^2=100\left(cm\right)\)

Vì 100cm=100cm

\(\Rightarrow BC^2=AB^2+AC^2\)

=> Tam giác ABC vuông tại A 

Xét diện tích tam giác ABC thường \(S_{ABCt}=\frac{AH.BC}{2}\left(1\right)\)

Xét diện tích tam giác ABC vuông \(S_{ABCv}=\frac{AC.AB}{2}\left(2\right)\)

Từ (1) và (2) 

\(\Leftrightarrow AH.BC=AB.AC\)

\(\Leftrightarrow AH.10=8.6\Leftrightarrow AH=4,8\left(cm\right)\)

Xét tam giác ABH vuông tại H 

\(\Rightarrow BH^2=AB^2-AH^2\left(PYTAGO\right)\)

\(\Rightarrow BH=\sqrt{AB^2-AH^2}\)

\(\Rightarrow BH=\sqrt{6^2-13,3^2}=3,6\left(cm\right)\)

Xét tam giác ACH vuông tại H

\(\Rightarrow HC^2=AC^2-AH^2\left(PYTAGO\right)\)

\(\Rightarrow HC=\sqrt{AC^2-AH^2}\)

\(\Rightarrow HC=\sqrt{8^2-4,8^2}=6,4\left(cm\right)\)

bút chì đọc tiếng anh là gì ?

Từ \(\frac{2019a+2020c}{2019a-2021c}=\frac{2019b+2020d}{2019b-2021d}\)

<=> \(\frac{2019a-2021c+4041c}{2019a-2021c}=\frac{2019b-2021d+4041d}{2019b-2021d}\)

<=> \(1+\frac{4041c}{2019a-2021c}=1+\frac{4041d}{2019b-2021d}\)

<=> \(\frac{4041c}{2019a-2021c}=\frac{4041d}{2019b-2021d}\)

<=> 4041c( 2019b - 2021d ) = 4041d( 2019a - 2021c )

<=> c( 2019b - 2021d ) = d( 2019a - 2021c )

<=> 2019bc - 2021dc = 2019ad - 2021cd

<=> 2019bc - 2021dc - 2019ad + 2021cd = 0

<=> 2019( bc - ad ) = 0

<=> bc - ad = 0

<=> bc = ad

<=> a/b = c/d

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

Ta có : \(\left(\frac{a+b}{c+d}\right)^{2020}=\left(\frac{kb+b}{kd+d}\right)^{2020}=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2020}=\left(\frac{b}{d}\right)^{2020}=\frac{b^{2020}}{d^{2020}}\)(1)

\(\)\(\frac{a^{2020}+b^{2020}}{c^{2020}+d^{2020}}=\frac{\left(kb\right)^{2020}+b^{2020}}{\left(kd\right)^{2020}+d^{2020}}=\frac{k^{2020}b^{2020}+b^{2020}}{k^{2020}d^{2020}+d^{2020}}=\frac{b^{2020}\left(k^{2020}+1\right)}{d^{2020}\left(k^{2020}+1\right)}=\frac{b^{2020}}{d^{2020}}\)(2)

Từ (1) và (2) ta có đpcm

\(\frac{13}{40}-\frac{11}{47}.\frac{1}{2}+\frac{1}{20}-\)\(\frac{121}{47.72}\)

\(=\left(\frac{13}{40}+\frac{1}{20}\right)-\)\(\left(\frac{11}{47}.\frac{1}{2}+\frac{121}{47.72}\right)\)

\(=\left(\frac{13}{40}+\frac{2}{40}\right)\)\(-\left(\frac{11}{47}.\frac{1}{2}+\frac{11}{47}.\frac{11}{72}\right)\)

\(=\frac{3}{8}-\frac{11}{47}.\left(\frac{1}{2}+\frac{11}{72}\right)\)

\(=\frac{3}{8}-\frac{11}{47}.\frac{47}{72}\)

\(=\frac{3}{8}-\frac{11}{72}\)

\(=\frac{27}{72}-\frac{11}{72}\)

\(=\frac{2}{9}\)

1.     A. headache      B. architect     C. chemical          D. children 

2.     A. although      B. enough       C. paragraph       D. cough

3.     A. heard            B. pearl            C. heart              D. earth

4.     A. put               B. junk             C. adult                D. sun

5.     A. future            B. circus          C. button              D. suggest

6.     A. church           B. amchair      C. architect         D. children

7.     A. banana          B. sofa            C. away               D. occasion

Ta có : \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)=))  \(x=2k;y=5k;z=7k\)

Thay vào biểu thức A ta được : 

\(A=\frac{2k-5k+7z}{2k+2.5k-7k}=\frac{4k}{5k}=\frac{4}{5}\)

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{x-y+z}{2-5+7}=\frac{x-y+z}{4}\) (1)

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{x}{2}=\frac{2y}{10}=\frac{z}{7}=\frac{x+2y-z}{2+10-7}=\frac{x+2y-z}{5}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{x-y+z}{4}=\frac{x+2y-z}{5}\Rightarrow A=\frac{x-y+x}{x+2y-z}=\frac{4}{5}\)