Ta có :

\(\frac{x-3}{97}+\frac{x-27}{73}+\frac{x-67}{33}+\frac{x-73}{27}=4\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}>0\) Nên \(x-100=0\)

\(\Leftrightarrow x=100\)

Vậy \(x=100\)

\(\Leftrightarrow\frac{x-3}{87}+\frac{x-27}{79}+\frac{x-67}{33}+\frac{x-73}{27}-4=0\)

\(\Leftrightarrow\left(\frac{x-3}{97}-1\right)+\left(\frac{x-27}{73}-1\right)+\left(\frac{x-67}{33}-1\right)+\left(\frac{x-73}{27}-1\right)=0\)

\(\Leftrightarrow\left(\frac{x-3-97}{97}\right)+\left(\frac{x-27-73}{73}\right)+\left(\frac{x-67-33}{33}\right)+\left(\frac{x-73-27}{27}\right)=0\)

\(\Leftrightarrow\frac{x-100}{97}+\frac{x-100}{73}+\frac{x-100}{33}+\frac{x-100}{27}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\right)=0\)

Vì \(\frac{1}{97}+\frac{1}{73}+\frac{1}{33}+\frac{1}{27}\ne0\)

\(\Rightarrow x-100=0\Leftrightarrow x=100\)

Bài 2

5 C

Bài 3

1 D

6 C

Còn lại ol r nhé

2) 5. C

3) 2. D

6. C

Còn lại ok nha

d. \(\dfrac{\pi}{2}< a;b< \pi\Rightarrow sina>0;sinb>0\)

\(sina=\sqrt{1-cos^2a}=\dfrac{4}{5}\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{4}{3}\)

\(sinb=\sqrt{1-cos^2b}=\dfrac{5}{13}\Rightarrow tanb=-\dfrac{5}{12}\)

Vậy:

\(sin\left(a-b\right)=sina.cosb-cosa.sinb=\dfrac{4}{5}.\left(-\dfrac{12}{13}\right)-\left(-\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)=...\)

\(cos\left(a-b\right)=cosa.cosb-sina.sinb=...\) (bạn tự thay số bấm máy)

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=...\)

\(cot\left(a+b\right)=\dfrac{1}{tan\left(a+b\right)}=\dfrac{1-tana.tanb}{tana+tanb}=...\)

e.

\(0< y< \dfrac{\pi}{2}\Rightarrow cosy>0\Rightarrow cosy=\sqrt{1-sin^2y}=\dfrac{4}{5}\)

\(\Rightarrow tany=\dfrac{siny}{cosy}=\dfrac{3}{4}\)

Vậy: \(tan\left(x+y\right)=\dfrac{tanx+tany}{1-tanx.tany}=...\)

\(cot\left(x-y\right)=\dfrac{1}{tan\left(x-y\right)}=\dfrac{1+tanx.tany}{tanx-tany}=...\)

Bài 3:

H đối xứng B qua AD

=>AD là đường trung trực của HB

=>AD⊥HB tại trung điểm của HB

mà AD⊥ AB

và HB,AB có điểm chung là B

nên B,A,H thẳng hàng

=>A là trung điểm của HB

Xét ΔIAH vuông tại A và ΔIAB vuông tại A có

IA chung

AH=AB

Do đó: ΔIAH=ΔIAB

=>\(\hat{AIH}=\hat{AIB}\)

mà \(\hat{AIH}=\hat{DIC}\) (hai góc đối đỉnh)

nên \(\hat{AIB}=\hat{DIC}\)

4 If you aren't alert, you computer will be infected with viruses

5 If we use much pesticide on vegetables, they will become poisonous

9 Because Nam needs to improve his social skills, he attends the activity

11 How about using public buses instead of motorbike 

12 They enjoy protecting the environment unpolluted 

15 Mr Brown was disappointed that his son didn't write to him

cái speed này đu hong kịp th đi coi phim =)))))

Bài 11:

a: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)

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

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

b: Để P=2 thì \(x-\sqrt{x}-2=0\)

hay x=4

Bài 10:

a: Ta có: \(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

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

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

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

b: Để A<0 thì \(\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Để A=-1 thì \(x+\sqrt{x}+1=-\sqrt{x}+1\)

\(\Leftrightarrow x=0\)

c: Thay x=4 vào A, ta được:

\(A=\dfrac{4+2+1}{2-1}=7\)