1:

(SAB), (SBC) vuông góc (BAC)

=>SB vuông góc (ABC)

AC vuông góc AB,SB

=>AC vuông góc (SAB)

=>AC vuông góc BH

mà SA vuông góc BH

nên BH vuông góc (SAC)

=>BH vuông góc SC

mà SC vuông góc BK

nên SC vuông góc (BHK)

c: (SH;(BHK))=góc SHK=(SA;BHK)

BC=BA/cos60=2a

SC=căn SB^2+BC^2=ăcn 5

SB^2=SK*SC

=>SK=a*căn 5/5

SA=căn SB^2+AB^2=a*căn 2

SB^2=SH*SA

=>SH=a*căn 2/2

sin SHK=căn 10/5

=>góc SHK=39 độ

S=a(a+2)

b: \(BC=\sqrt{89}\left(cm\right)\)

\(\sin\widehat{B}=\dfrac{5\sqrt{89}}{89}\)

\(\Leftrightarrow\widehat{B}\simeq32^0\)

\(\widehat{C}=58^0\)

Bài 1.

a, PTPƯ: kẽm + axit clohidric → kẽm clorua + hidro

b, Theo ĐLBTKL ta có:

 \(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)

c, Ta có: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=27,2+0,4-13=14,6\left(g\right)\)

Bài 2:

a, PTPƯ: metan + oxi → cacbon dioxit + hơi nước

b, Theo ĐLBTKL ta có: 

 \(m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\)

c, Ta có: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=132+108-48=192\left(g\right)\)  

 PT chữ: Metan + Oxi \(\xrightarrow[]{t^o}\) Cacbon đioxit + Nước

Bảo toàn khối lượng: \(m_{O_2}=m_{CO_2}+m_{H_2O}-m_{CH_4}=192\left(g\right)\)

Bài 1:

PT chữ: Kẽm + Axit clohidric \(--->\) Kẽm clorua + Hidro

Bảo toàn khối lượng: \(m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=14,6\left(g\right)\)

\(b,\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{\sqrt{15}}=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(d,\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\left(\sqrt{ab}-\sqrt{bc}\right)}=\sqrt{ab}+\sqrt{bc}=\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)\)

\(e,\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\left(\sqrt{\dfrac{a}{b}+\dfrac{2b.\sqrt{ab}}{b}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)

\(=a\sqrt{a}\sqrt{a+2b\sqrt{ab}}+b\sqrt{a^2}\)

\(=a\sqrt{a^2+2ab\sqrt{ab}}+ab\)

\(=a\left(\sqrt{a^2+2ab\sqrt{ab}}+b\right)\)

\(f,\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)

\(=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\)

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

\(=\left(a-1\right)^2=a^2-2a+1\)

\(d,=\dfrac{3y}{5x\left(x-y\right)}\\ e,=\dfrac{5x\left(x+2\right)\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\dfrac{-5x}{4}\\ f,=\dfrac{3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(6-x\right)}=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\\ g,=\dfrac{3xy\left(x-3y\right)\left(x+3y\right)}{2x^2y^2\left(x-3y\right)}=\dfrac{3\left(x+3y\right)}{2xy}\\ h,=\dfrac{45x^2y\left(x-y\right)\left(x+y\right)}{10xy\left(y-x\right)}=\dfrac{-9x\left(x+y\right)}{2}\\ i,=\dfrac{12\left(a-b\right)\left(a+b\right)\left(a^2+ab+b^2\right)}{3\left(a+b\right)\left(a-b\right)^2}=\dfrac{4\left(a^2+ab+b^2\right)}{a-b}\)

e: \(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=-\dfrac{5}{2}\)

1.

c, \(sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\\dfrac{\pi}{3}-x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}-arcsin\left(-\dfrac{1}{4}\right)+k.360^o\\x=-\dfrac{2\pi}{3}+arcsin\left(-\dfrac{1}{4}\right)+k.360^o\end{matrix}\right.\)

d, \(sin4x=\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=arcsin\dfrac{2}{3}+k2\pi\\4x=\pi-arcsin\dfrac{2}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}arcsin\dfrac{2}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arcsin\dfrac{2}{3}+\dfrac{k\pi}{2}\end{matrix}\right.\)

1.

e, \(2sin2x+\sqrt{2}=0\)

\(\Leftrightarrow sin2x=-\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x=sin\left(-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)

1.

f, \(sin\left(3x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=x-\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{4}=-x+\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{24}+k\pi\\x=\dfrac{11\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Lỗi ảnh r 

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