Bài 3: 

a: Ta có: \(x-2\sqrt{x+8}=0\)

\(\Leftrightarrow\sqrt{4x+32}=x\)

\(\Leftrightarrow4x+32=x^2\)

\(\Leftrightarrow x^2-4x-32=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

b: Ta có: \(3\sqrt{x+5}=4x-7\)

\(\Leftrightarrow\left(4x-7\right)^2=9x+45\)

\(\Leftrightarrow16x^2-56x+49-9x-45=0\)

\(\Leftrightarrow16x^2-65x+4=0\)

\(\Leftrightarrow16x^2-64x-x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(16x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{16}\end{matrix}\right.\)

Câu 2: 

a: f(-1)=-2

f(0)=0

f(2)=4

1: 

#include <bits/stdc++.h>

using namespace std;

long long x,i,n,t;

int main()

{

cin>>n;

t=0;

for (i=1; i<=n; i++)

{

cin>>x;

if (x%2==0) t=t+x;

}

cout<<t;

return 0;

}

2: 

#include <bits/stdc++.h>

using namespace std;

long long x,n,i,dem;

int main()

{

cin>>n;

dem=0;

for (i=1; i<=n; i++)

{

cin>>x;

if (x<0) dem++;

}

cout<<dem;

return 0;

}

Ơ đây là bài lớp mấy z

1.b

2d

3.a

4.c

5.c

6. b

7.a

8.d

9.d

10.d

cong ở 4 chỗ là sao? Mik khum hỉu cho lắm

a: Xét ΔAHB vuông tại H và ΔCAB vuông tại A có 

góc B chung

Do đó: ΔAHB\(\sim\)ΔCAB

Suy ra: BA/BC=BH/BA

hay \(BA^2=BH\cdot BC\)

b: \(AC=\sqrt{BC^2-AB^2}=24\left(cm\right)\)

\(HC=\dfrac{AC^2}{BC}=\dfrac{24^2}{40}=14.4\left(cm\right)\)

Lời giải:

Nếu $x+y+z+t=0$ thì $M=\frac{-t}{t}=\frac{-x}{x}=\frac{-z}{z}=-1$

$\Rightarrow (M-1)^{2025}=(-1-1)^{2025}=(-2)^{2025}$

Nếu $x+y+z+t\neq 0$. Áp dụng TCDTSBN:

$M=\frac{x+y+z}{t}=\frac{y+z+t}{x}=\frac{z+t+x}{y}=\frac{t+x+y}{z}=\frac{x+y+z+y+z+t+z+t+x+t+x+y}{t+x+y+z}=\frac{3(x+y+z+t)}{x+y+z+t}=3$

$\Rightarrow (M-1)^{2025}=2^{2025}$

1. you be quiet, you'll hear the sound.

2. I am short, I can't become a model.

3. you don't run quickly, you can't catch the bus.

4. Jack doesn't pass the exam, his parents will be sad.

5. you don't be hurry, you'll be late.

6. they don't tell the truth, we can't know about this.

7. you speak slowly, I'll understand you.

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)