\(a.\frac{19}{13}+\frac{-7}{11}+\frac{-4}{11}-\frac{6}{13}\)

\(=\left(\frac{19}{13}-\frac{6}{13}\right)+\left(\frac{-7}{11}+\frac{-4}{11}\right)\)

\(=1+\left(-1\right)\)

\(=0\)

(19/13-6/13)+(-7/11+-4/11)

1+-1

0

1/2:1/2-1/4:1/4+1/8:5/4-1/10:1/10

1/2.2/1-1/4.4/1+1/8.4/5-1/10.1/10

1-1+1/8.4/5-1

1/10

ac=4 b)

ac là cạnh chung

ab=ad

dac=bac

biết tới đây thui :(( sorry

$\frac{999}{1000}<\frac{1000}{1001}$

$-\frac{15}{32}<-\frac{63}{128}$

a)\(\frac{999}{1000}< \frac{1000}{1001}\)                                     \(-\frac{15}{32}< -\frac{63}{128}\)

đề bài có đúng ko đấy

\(A\left(x\right)=2x^2+2x+3\)

3) \(A\left(x\right)=3\)

khi đó: \(2x^2+2x+3=3\)

<=> \(x^2+x=0\)

<=> \(x\left(x+1\right)=0\)

<=> \(x=0\)

hoặc \(x=-1\)

A(x) = 3x² + x³ + 5x⁴ - x² - x³ - 5x⁴ + 2x + 3 

        = 2x² + 2x + 3

A(x) + B(x) = 2x - 7

<=> ( 2x² + 2x + 3 ) + B(x) = 2x - 7

B(x) = 2x - 7 - ( 2x² + 2x + 3 )

        = 2x - 7 - 2x² - 2x - 3

        = -2x² - 10

A(x) = 3 <=> 2x² + 2x + 3 = 3

              <=> x( 2x + 2 ) = 0

              <=> x = 0 hoặc 2x + 2 = 0

              <=> x = 0 hoặc x = -1 

a) 

Xét \(\Delta\)HBA và \(\Delta\)HAC 

có: ^BHA = ^AHC = 90 độ 

^HBA = ^HAC ( cùng phụ ^HAB ) 

=> \(\Delta\)HBA ~ \(\Delta\)HAC 

b) Ta có: \(BC=\sqrt{AB^2+AC^2}=10\)cm

=> \(S\left(ABC\right)=\frac{1}{2}AB.AC=\frac{1}{2}AH.BC\)

=> \(AH=\frac{6.8}{10}=4,8\)cm

c) Tích chất phân giác

=> \(\frac{AB}{BC}=\frac{AD}{DC}\Rightarrow\frac{AD}{6}=\frac{DC}{10}=\frac{AD+DC}{6+10}=\frac{8}{16}=\frac{1}{2}\)

=> AD = 3 cm; DC = 5 cm 

Theo pi ta go trong \(\Delta\)ADB => \(BD=\sqrt{AB^2+AD^2}=\sqrt{6^2+3^2}=3\sqrt{5}\)

                                                A B C D H

a) \(\Delta ABC\)vuông tại A \(\Rightarrow\widehat{ABC}+\widehat{C}=90^o\)

\(\Delta AHC\)vuông tại H \(\Rightarrow\widehat{HAC}+\widehat{C}=90^o\)

\(\Rightarrow\widehat{HAC}=\widehat{ABC}\)

Xét \(\Delta HBA\)và \(\Delta HAC\)có:+) \(\widehat{AHB}=\widehat{AHC}=90^o\)

                                                    +) \(\widehat{HAC}=\widehat{ABC}\)

\(\Rightarrow\Delta HBA~\Delta HAC\left(g-g\right)\)( đpcm )

b) \(\Delta ABC\)vuông tại A \(\Rightarrow AB^2+AC^2=BC^2\)( định lý Pytago )

\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\)

Xét \(\Delta ABC\)có: \(S=\frac{1}{2}AB.AC=\frac{1}{2}AH.BC\)

\(\Rightarrow AB.AC=AH.BC\)\(\Rightarrow AH=\frac{AB.AC}{BC}=\frac{6.8}{10}=4,8\)

c) \(\Delta ABC\)có BD là phân giác \(\Rightarrow\frac{AB}{BC}=\frac{AD}{DC}=\frac{6}{10}=\frac{3}{5}\)

\(\Rightarrow\frac{AD}{3}=\frac{DC}{5}=\frac{AD+DC}{3+5}=\frac{AC}{8}=\frac{8}{8}=1\)

\(\Rightarrow DC=5.1=5\); \(AD=3.1=3\)

Xét \(\Delta ABD\)vuông tại A \(\Rightarrow AB^2+AD^2=BD^2\)( định lý Pytago )

\(\Rightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{6^2+3^2}=\sqrt{54}=3\sqrt{6}\)

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\(M=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4005}\)

\(\frac{M}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{8010}\)

\(\frac{M}{2}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{89x90}\)

\(\frac{M}{2}=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{90-89}{89.90}\)

\(\frac{M}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{89}-\frac{1}{90}=\frac{1}{3}-\frac{1}{90}\)

\(M=\frac{2}{3}-\frac{2}{90}< \frac{2}{3}\)

A) 9 giờ

B) 19 giờ 30 phút.