A B C H K E

d) Ta có:

\(AC^2=HC.BC\)

\(\Rightarrow8^2=HC.10\)(thay số).

\(\Rightarrow10HC=64\Rightarrow HC=6,4\left(dm\right)\)

Xét \(\Delta HCE\)và \(\Delta ACK\)có:

\(\widehat{CHE}=\widehat{CAK}\left(=90^0\right)\)

\(\widehat{HCE}=\widehat{ACK}\)(vì \(CK\)là phân giác của \(\widehat{ACB}\))

\(\Rightarrow\Delta HCE~\Delta ACK\left(g.g\right)\)

\(\Rightarrow\frac{CH}{CA}=\frac{EH}{KA}\)(2 cặp cạnh tương ứng tỉ lệ).

\(\Rightarrow\frac{6,4}{8}=\frac{4}{5}=\frac{EH}{KA}\)(thay số).

Do đó \(\frac{S_{HCE}}{S_{ACK}}=\left(\frac{EH}{KA}\right)^2=\left(\frac{4}{5}\right)^2=\frac{16}{25}\left(1\right)\)

Bạn hãy chỉ ra \(KA=\frac{8}{3}dm\)

Rồi hãy chỉ ra \(S_{AKC}=\frac{\frac{8}{3}.8}{2}=\frac{32}{3}\left(dm^2\right)\left(2\right)\)

Thay (2) vào (1), ta được:

\(\frac{S_{HCE}}{\frac{32}{3}}=\frac{16}{25}\Rightarrow S_{HCE}=\frac{16}{25}.\frac{32}{3}=\frac{512}{75}\left(dm^2\right)\)

Vậy \(S_{HCE}=\frac{512}{75}dm^2\)

Xem tớ có sai sót chỗ nào không?

ĐKXĐ : x khác 0 ; x khác 5

<=> \(\frac{45\left(x-5\right)}{x\left(x-5\right)}-\frac{45x}{x\left(x-5\right)}=\frac{3}{2}\)

<=> \(\frac{-225}{x\left(x-5\right)}=\frac{3}{2}\)

=> 3x( x - 5 ) = -450

<=> 3x² - 15x + 450 = 0

<=> x² - 5x + 150 = 0

Vì x² - 5x + 150 = ( x - 5/2 )² + 575/4 ≥ 575/4 ∀ x

nên pt vô nghiệm

ĐKXĐ : x khác 0 ; x khác 5

<=> \(\frac{45\left(x-5\right)}{x\left(x-5\right)}-\frac{45x}{x\left(x-5\right)}=\frac{3}{2}\)

<=> \(\frac{-225}{x\left(x-5\right)}=\frac{3}{2}\)

=> 3x( x - 5 ) = -450

<=> 3x² - 15x + 450 = 0

<=> x² - 5x + 150 = 0

Vì x² - 5x + 150 = ( x - 5/2 )² + 575/4 ≥ 575/4 ∀ x

nên pt vô nghiệm

Trả lời:

2x - ( 3 - 5x ) = 4 ( x + 3 )

<=> 2x - 3 + 5x = 4x + 12

<=> 7x - 3 = 4x + 12

<=> 7x - 4x = 12 + 3

<=> 3x = 15

<=> x = 5

Vậy S = { 5 }

ĐKXĐ : x ≥ -1/2

Bình phương hai vế

=> x⁴ - 2x² + 1 = 4x² + 4x + 1

<=> x⁴ - 6x² - 4x = 0

<=> x( x³ - 6x - 4 ) = 0

<=> x( x³ + 2x² - 2x² - 4x - 2x - 4 ) = 0

<=> x[ x²( x + 2 ) - 2x( x + 2 ) - 2( x + 2 ) ] = 0

<=> x( x + 2 )( x² - 2x - 2 ) = 0

<=> x( x + 2 )( x - 1 - √3 )( x - 1 + √3 ) = 0

<=> x = 0 hoặc x = -2 hoặc x = 1 ± √3

Đối chiếu với ĐKXĐ ta thấy x = 0 và x = 1 + √3 thỏa mãn

Vậy phương trình có tập nghiệm S = { 0 ; 1 + √3 }

ĐKXĐ : \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

a) \(\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\)

<=> \(\frac{1}{4}x-\frac{5}{4}-2x+1=\frac{1}{3}x-\frac{1}{3}+\frac{1}{6}x\)

<=> \(-\frac{7}{4}x-\frac{1}{2}x=-\frac{1}{3}+\frac{1}{4}\)

<=> \(-\frac{9}{4}x=-\frac{1}{12}\)

<=> \(x=\frac{1}{27}\)

Vậy ...

b) ( x² - 4 ) - ( x - 2 )( 3 - 2x ) = 0

<=> ( x - 2 )( x + 2 ) - ( x - 2 )( 3 - 2x ) = 0

<=> ( x - 2 )( x + 2 - 3 + 2x ) = 0

<=> ( x - 2 )( 3x - 1 ) = 0

<=> x = 2 hoặc x = 1/3

Vậy ...

a) \(\frac{x-3}{5}+\frac{1+2x}{3}=6\)

<=> \(\frac{1}{5}x-\frac{3}{5}+\frac{1}{3}+\frac{2}{3}x=6\)

<=> \(\frac{13}{15}x=\frac{94}{15}\)<=> \(x=\frac{94}{13}\)

Vậy pt có nghiệm x = 94/13

b) ( 2x - 3 )( x² + 1 ) = 0 (1)

Vì x² + 1 ≥ 1 > 0 ∀ x

nên (1) <=> 2x - 3 = 0 <=> x = 3/2

Vậy pt có nghiệm x = 3/2

c) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\right)\)

<=> \(\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x-2\right)}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

=> 2x - 4 - x - 1 = 3x - 11

<=> -4x = -6 <=> x = 3/2 (tm)

Vậy pt có nghiệm x = 3/2

a)\(\frac{x-3}{5}+\frac{1+2x}{3}=6\Leftrightarrow6\left(x-3\right)+10\left(1+2x\right)=180\)

\(\Leftrightarrow6x-18+10+20x=180\)

\(\Leftrightarrow26x-8=180\)

\(\Leftrightarrow26x=188\Rightarrow x=\frac{188}{26}=\frac{94}{13}\)

b)\(\left(2x-3\right)\left(x^2+1\right)=0\)

\(\orbr{\begin{cases}2x-3=0\\x^2+1=0\end{cases}\Rightarrow x=\frac{3}{2}}\)

c)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)-1\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow11-5=3x-x\)

\(\Leftrightarrow6=2x\Rightarrow x=3\)