3. I think collecting stamps is interesting

Giúp e 2 câu còn lại với ạ

Question 2: David has volunteered for 2 years

Question 3: I think collecting stamps is interesting

chị đc giáo viên nào like nhìu nhất zậy

mình làm câu 2, 3 rồi nhé

https://hoc24.vn/cau-hoi/anh-chi-giup-em-tieng-anh-7-cau-1-va-2-voi-a-chieu-e-thi-aem-cam-on-anh-chi-rat-nhieu-a.3054793966970

câu 1: Lan had a high fever, so she stayed home from school yesterday

h=1,4+tan39o.400 ≈325(m)

Bài 13:

ĐKXĐ: x∉{0;2;-2;1/2}

a: \(B=\left(\frac{x+2}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{2x^2-x}{x^2-2x}\)

\(=\left(\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(2x-1\right)}{x\left(x-2\right)}\)

\(=\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x-1}\)

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{x+2}\cdot\frac{1}{2x-1}=\frac{-4x^2-8x}{\left(x+2\right)\left(2x-1\right)}\)

\(=\frac{-4x\left(x+2\right)}{\left.\left(x+2\right)\left(2x-1\right)\right.}=\frac{-4x}{2x-1}\)
b: |x|=3

=>x=3 hoặc x=-3

Khi x=3 thì \(B=\frac{-4\cdot3}{2\cdot3-1}=\frac{-12}{5}\)

Khi x=-3 thì \(B=\frac{-4\cdot\left(-3\right)}{2\cdot\left(-3\right)-1}=\frac{12}{-6-1}=\frac{-12}{7}\)

c: Để B nguyên thì -4x⋮2x-1

=>-4x+2-2⋮2x-1

=>-2⋮2x-1

mà 2x-1 lẻ

nên 2x-1∈{1;-1}

=>2x∈{2;0}

=>x∈{1;0}

Kết hợp ĐKXĐ, ta được: x=1

Bài 12:

a: ĐKXĐ: a∉{1;-1;-2}

b: \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right)\cdot\frac{2a+2}{a+2}\)

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

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

c: |a|=2

=>a=2(nhận) hoặc a=-2(loại)

Khi a=2 thì \(P=\frac{2}{2-1}=\frac21=2\)

Bài 11:

a: ĐKXĐ: x∉{2;-3}

b: \(P=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

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

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

c: \(P=\frac{-3}{4}\)

=>\(\frac{x-4}{x-2}=\frac{-3}{4}\)

=>4(x-4)=-3(x-2)

=>4x-16=-3x+6

=>7x=22

=>\(x=\frac{22}{7}\) (nhận)

d: Để P nguyên thì x-4⋮x-2

=>x-2-2⋮x-2

=>-2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

e: \(x^2-9=0\)

=>\(x^2=9\)

=>x=3(nhận) hoặc x=-3(loại)

Khi x=3 thì \(P=\frac{3-4}{3-2}=-1\)

a: \(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(=\dfrac{2\sqrt{x}+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}}{x-1}\)

\(P=\left(\dfrac{2}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{x-1}\)

\(\Rightarrow P=\dfrac{\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}-1}\)

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{2+2\sqrt{2}}\)

\(\Rightarrow P=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}\)

\(\Rightarrow P=\dfrac{1}{2}\)

a: \(\frac{2x}{3}:\frac{5}{6x^2}=\frac{2x}{3}\cdot\frac{6x^2}{5}=\frac{12x^3}{15}=\frac{4x^3}{5}\)

b: \(16x^2y^2:\left(-\frac{18x^2y^5}{5}\right)\)

\(=16x^2y^2\cdot\frac{-5}{18x^2y^5}=\frac{-80x^2y^2}{18x^2y^5}=\frac{-40}{9y^3}\)

c: \(\frac{25x^3y^5}{3}:15xy^2=\frac{25x^3y^5}{3\cdot15xy^2}=\frac{25x^3y^5}{45xy^2}=\frac59x^2y^3\)

d: \(\frac{x^2-y^2}{6x^2y}:\frac{x+y}{3xy}=\frac{\left(x-y\right)\left(x+y\right)}{6x^2y}\cdot\frac{3xy}{x+y}=\frac{x-y}{2x}\)

e: \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}\)

\(=\frac{a\left(a+b\right)}{b-a}\cdot\frac{2\left(a^2-b^2\right)}{a+b}=\frac{a\cdot2\cdot\left(a-b\right)\left(a+b\right)}{b-a}=-2a\left(a+b\right)\)

f: \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}\)

\(=\frac{-\left(x+y\right)}{x-y}\cdot\frac{3\left(x^2-y^2\right)}{x\left(x+y\right)}=\frac{-3\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)}=\frac{-3\cdot\left(x+y\right)}{x}\)

g: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)

h: \(\frac{5x-15}{4x+4}:\frac{x^2-9}{x^2+2x+1}\)

\(=\frac{5\left(x-3\right)}{4\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{5\left(x+1\right)}{4\left(x+3\right)}\)

i: \(\frac{6x+48}{7x-7}:\frac{x^2-64}{x^2-2x+1}\)

\(=\frac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\frac{6\left(x-1\right)}{7\left(x-8\right)}\)

k: \(\frac{4x-24}{5x+5}:\frac{x^2-36}{x^2+2x+1}\)

\(=\frac{4\left(x-6\right)}{5\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-6\right)\left(x+6\right)}=\frac{4\left(x+1\right)}{5\left(x+6\right)}\)

l: \(\frac{3x+21}{5x+5}:\frac{x^2-49}{x^2+2x+1}\)

\(=\frac{3\left(x+7\right)}{5\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-7\right)\left(x+7\right)}=\frac{3\left(x+1\right)}{5\left(x-7\right)}\)

m: \(\frac{3-3x}{\left(1+x\right)^2}:\frac{6x^2-6}{x+1}\)

\(=\frac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\frac{x+1}{6\left(x-1\right)\left(x+1\right)}=\frac{-3}{6\left(x+1\right)^2}=\frac{-1}{2\left(x+1\right)^2}\)

ΔABC vuông tại A

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

=>\(BC^2=8^2+5^2=64+25=89\)

=>\(BC=\sqrt{89}\) ≃9,4(cm)

Xét ΔABC vuông tại A có tan B=\(\frac{AC}{AB}=\frac58\)

nên \(\hat{B}\) ≃32 độ

ΔABC vuông tại A

=>\(\hat{ABC}+\hat{ACB}=90^0\)

=>\(\hat{ACB}=90^0-32^0=58^0\)

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì?