ΔAHB vuông tại H

=>\(AB^2=AH^2+HB^2\)

=>\(HB=\sqrt{15^2-12^2}=9\left(cm\right)\)

ΔAHC vuông tại H

=>\(AH^2+HC^2=AC^2\)

=>\(HC=\sqrt{20^2-12^2}=16\left(cm\right)\)

BC=BH+CH

=9+16

=25(cm)

AH là đường cao ⇒ AH ⊥ BC 

Áp dụng đính lý Py-ta-go cho ΔAHB vuông tại H ta có:

\(AH^2+BH^2=AB^2\)

\(\Rightarrow BH=\sqrt{AB^2-AH^2}=\sqrt{15^2-12^2}=9\)

Áp dụng định lý Py-ta-go cho ΔAHC vuông tại H ta có: 

\(AH^2+CH^2=AC^2\)

\(\Rightarrow CH=\sqrt{AC^2-AH^2}=\sqrt{20^2-12^2}=16\)

Mà: \(BC=BH++CH\Rightarrow BC=9+16=25\)

\(\dfrac{7^{49}-7^{48}}{7^{48}}\)

\(=\dfrac{7^{49}}{7^{48}}-\dfrac{7^{48}}{7^{48}}\)

\(=7-1\)

\(=6\)

\(\dfrac{7^{49}-7^{48}}{7^{48}}\)

\(=\dfrac{7^{48}\cdot7-7^{48}}{7^{48}}\)

\(=\dfrac{7^{48}\left(7-1\right)}{7^{48}}\)

\(=7-1\)

\(=6\)

13) 

a) \(\left\{{}\begin{matrix}7x+4y=2\\5x-2y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x+4y=2\\10x-4y=32\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x+4y=2\\17x=34\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\cdot2+4y=2\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4y=2-14\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=-12\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

Vậy: .... 

b) \(\left\{{}\begin{matrix}2x+3y=19\\3x+4y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+9y=57\\6x+8y=-28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=19\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot85=19\\y=85\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=19-255\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-236\\y=85\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-118\\y=85\end{matrix}\right.\)

Vậy: .... 

c) \(\left\{{}\begin{matrix}2x+2y=3\\3x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\3x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3\cdot1-2y=2\end{matrix}\right.\)

 \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\-2y=2-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: ....

15) 

a) \(\left\{{}\begin{matrix}5\left(x+2\right)=2\left(y+7\right)\\3\left(x+y\right)=17-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+10=2y+14\\3x+3y=17-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=14-10\\3x+3y+x=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=4\\4x+3y=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=12\\8x+6y=34\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=4\\23x=46\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\cdot2-2y=4\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=6\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

vậy: ... 

wow

Mik k nghĩ Gia Huy là tên con gái á bnn ^^

a: Thay x=0 vào A, ta được:

\(A=\dfrac{0+1}{0-3}=\dfrac{1}{-3}=-\dfrac{1}{3}\)

b: \(B=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

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

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

\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)

c: \(P=\dfrac{B}{A}=\dfrac{3x}{x-3}:\dfrac{x+1}{x-3}=\dfrac{3x}{x+1}\)

Để P nguyên thì \(3x⋮x+1\)

=>\(3x+3-3⋮x+1\)

=>\(-3⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

`#3107.101107`

`a.`

Tại `x = 0:`

`A = (0 + 1)/(0 - 3) = 1/(-3) = -1/3`

Vậy, `A = -1/3` tại `x = 0`

`b.`

`B= (2x)/(x+3) + (x+1)/(x-3) + (3 - 11x)/(9-x^2)`

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

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

\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{\left(2x^2+x^2\right)+\left(11x-6x+4x\right)+\left(3-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{3x}{x-3}\)

`c.`

`P = B/A`

`=> P =` \(\dfrac{3x}{x-3}\div\dfrac{x+1}{x-3}\)

`P = (3x)/(x - 3) * (x - 3)/(x + 1)`

`P = (3x)/(x + 1)`

Ta có: `3x \vdots (x + 1)`

`=> (3x + 3 - 3) \vdots (x + 1)`

`=> -3 \vdots (x + 1)`

`=> x + 1 \in \text{Ư(-3)} = {+-1; +-3}`

`=>`\(\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=3\\x+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=2\\x=-4\end{matrix}\right.\)

Nghiệm của `x^2+10x+21` là:

\(x^2+10x+21=0\)

\(\Leftrightarrow x^2+3x+7x+21=0\)

\(\Leftrightarrow x\left(x+3\right)+7\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+7\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-7\end{matrix}\right.\)

Thay lần lượt `x=-3;x=-7` vào `(x+2)(x+4)(x+6)(x+8)+2017` ta có:

\(\left(-3+2\right)\left(-3+4\right)\left(-3+6\right)\left(-3+8\right)+2017=\left(-1\right)\cdot1\cdot3\cdot5=2002\) 

\(\left(-7+2\right)\left(-7+4\right)\left(-7+6\right)\left(-7+8\right)+2017=\left(-5\right)\cdot\left(-3\right)\cdot\left(-1\right)\cdot1+2017=2002\)

Vậy số dư của `(x+2)(x+4)(x+6)(x+8)+2017` khi chia cho `x^2+10x+21` là 2002 

Xem lại đề nhé em!

Gọi giá tiền mỗi kg gạo mà bác Lan đã mua là \(x\) (nghìn đồng; \(x>2\))

Số gạo bác Lan đã mua là: \(\dfrac{480}{x}\) (kg)

Giá tiền mỗi kg gạo theo dự định là: \(x-2\) (nghìn đồng)

Số gạo bác Lan dự định mua là: \(\dfrac{480}{x-2}\) (kg)

Vì bác Lan đã mua lượng gạo giảm \(\dfrac{1}{16}\) lần so với dự định nên ta có phương trình:

\(\dfrac{480}{x-2}.\left(1-\dfrac{1}{16}\right)=\dfrac{480}{x}\)

\(\Leftrightarrow\dfrac{15}{16\left(x-2\right)}=\dfrac{1}{x}\)

\(\Rightarrow15x=16x-32\)

\(\Leftrightarrow x=32\left(tm\right)\)

Vậy giá tiền mỗi kg gạo mà bác Lan đã mua là 32000 đồng.

a: \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

=>\(\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

=>\(\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

=>\(\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

=>x+100=0

=>x=-100

b: \(\dfrac{x-214}{86}+\dfrac{x-132}{84}+\dfrac{x-54}{82}=6\)

=>\(\left(\dfrac{x-214}{86}-1\right)+\left(\dfrac{x-132}{84}-2\right)+\left(\dfrac{x-54}{82}-3\right)=0\)

=>\(\dfrac{x-300}{86}+\dfrac{x-300}{84}+\dfrac{x-300}{82}=0\)

=>x-300=0

=>x=300

c: \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

=>\(\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

=>\(\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

=>123-x=0

=>x=123

d: \(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)

=>\(\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=0\)

=>\(\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)

=>x-2007=0

=>x=2007

a) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

\(\Leftrightarrow\dfrac{x+2}{98}+\dfrac{x+4}{96}-\dfrac{x+6}{94}-\dfrac{x+8}{92}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)-\left(\dfrac{x+6}{94}+1\right)-\left(\dfrac{x+8}{92}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\ne0\))

\(\Leftrightarrow x=-100\)

b) \(\dfrac{x-214}{86}+\dfrac{x-132}{84}+\dfrac{x-54}{82}=6\)

\(\Leftrightarrow\left(\dfrac{x-214}{86}-1\right)+\left(\dfrac{x-132}{84}-2\right)+\left(\dfrac{x-54}{82}-3\right)=0\)

\(\Leftrightarrow\dfrac{x-300}{86}+\dfrac{x-300}{84}+\dfrac{x-300}{82}=0\)

\(\Leftrightarrow\left(x-300\right)\left(\dfrac{1}{86}+\dfrac{1}{84}+\dfrac{1}{82}\right)=0\)

\(\Leftrightarrow x-300=0\) (vì \(\dfrac{1}{86}+\dfrac{1}{84}+\dfrac{1}{82}\ne0\))

\(\Leftrightarrow x=300\)

c) \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow123-x=0\) (vì \(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\))

\(\Leftrightarrow x=123\)

d) \(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)

\(\Leftrightarrow\left(\dfrac{x-17}{1990}-1\right)+\left(\dfrac{x-21}{1986}-1\right)+\left(\dfrac{x+1}{1004}-2\right)=0\)

\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)

\(\Leftrightarrow x-2007=0\) (vì \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))

\(\Leftrightarrow x=2007\)

$Toru$

\(3x^3-14x^2+4x+3\)

\(=3x^3+x^2-15x^2-5x+9x+3\)

\(=x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-5x+3\right)\)

\(\dfrac{4x^2+16}{x^2+6}=\dfrac{3}{x^2+1}+\dfrac{5}{x^2+3}+\dfrac{7}{x^2+5}\)

=>\(\dfrac{4x^2+16}{x^2+6}-3=\dfrac{3}{x^2+1}-1+\dfrac{5}{x^2+3}-1+\dfrac{7}{x^2+5}-1\)

=>\(\dfrac{x^2-2}{x^2+6}=\dfrac{-x^2+2}{x^2+1}+\dfrac{-x^2+2}{x^2+3}+\dfrac{-x^2+2}{x^2+5}\)

=>\(\dfrac{x^2-2}{x^2+6}+\dfrac{x^2-2}{x^2+1}+\dfrac{x^2-2}{x^2+3}+\dfrac{x^2-2}{x^2+5}=0\)

=>\(\left(x^2-2\right)\left(\dfrac{1}{x^2+6}+\dfrac{1}{x^2+1}+\dfrac{1}{x^2+3}+\dfrac{1}{x^2+5}\right)=0\)

=>\(x^2-2=0\)

=>\(x^2=2\)

=>\(x=\pm\sqrt{2}\)