x² + 2y² +3xy - x - y + 3 = 0

(x² - y²) + (3y² + 3xy) - (x + y) = -3

(x - y)(x + y) + 3y(x + y) - (x + y) = -3

(x + y)(x + 2y -1) = -3 = 1.(-3) = (-1).3

(x;y)=(4;-3) (-6;5)

Lời giải:

$2(6x+7)^2(3x+4)(x+1)-12=0$

$\Leftrightarrow 2(36x^2+84x+49)(3x^2+7x+4)-12=0$

Đặt $3x^2+7x+4=a$ thì PT trở thành:

$2(12a+1)a-12=0$

$\Leftrightarrow 2a(12a+1)-12=0$

$\Leftrightarrow 24a^2+2a-12=0$

$\Leftrightarrow (24a^2-16a)+(18a-12)=0$

$\Leftrightarrow 8a(3a-2)+6(3a-2)=0$

$\Leftrightarrow (3a-2)(8a+6)=0$

$\Leftrightarrow (3a-2).2(4a+3)=0$

$\Leftrightarrow (3a-2)(4a+3)=0$

$\Rightarrow 3a-2=0$ hoặc $4a+3=0$

Nếu $3a-2=0$

$\Leftrightarrow 3(3x^2+7x+4)-2=0$

$\Leftrightarrow 9x^2+21x+10=0$

$\Leftrightarrow (3x+2)(3x+5)=0\Leftrightarrow x=\frac{-2}{3}$ hoặc $x=\frac{-5}{3}$

Nếu $4a+3=0$

$\Leftrightarrow 4(3x^2+7x+4)+3=0$

$\Leftrightarrow 12x^2+28x+19=0$

$\Leftrightarrow 12(x+\frac{7}{6})^2=\frac{-8}{3}<0$ (vô lý - loại)

Vậy..........

a: Ta có: \(AE=EB=\dfrac{AB}{2}\)

\(BF=FC=\dfrac{BC}{2}\)

\(DK=KC=\dfrac{DC}{2}\)

mà AB=BC=CD

nên AE=EB=BF=FC=DK=KC

Xét tứ giác AECK có

AE//CK

AE=CK

Do đó: AECK là hình bình hành

b: Xét ΔDCF vuông tại C và ΔCBE vuông tại B có

DC=CB

CF=BE

Do đó: ΔDCF=ΔCBE

=>\(\widehat{DFC}=\widehat{CEB}\)

mà \(\widehat{CEB}+\widehat{BCE}=90^0\)

nên \(\widehat{BCE}+\widehat{DFC}=90^0\)

=>CE\(\perp\)DF

a) \(\dfrac{12}{5}>\dfrac{10}{5}=2=\dfrac{4}{2}>\dfrac{3}{2}\) (Số 2 làm trung gian)

Hay \(\dfrac{12}{5}>\dfrac{3}{2}\)

b) Ta có: 

`2023 < 2024 =>` \(\dfrac{2023}{2024}< 1\)

`2026 > 2025 =>` \(\dfrac{2026}{2025}>1\)

=> \(\dfrac{2023}{2024}< 1< \dfrac{2026}{2025}\) (1 làm trung gian)

Hay \(\dfrac{2023}{2024}< \dfrac{2026}{2025}\)

a: \(\dfrac{12}{5}=2,4;\dfrac{3}{2}=1,5\)

mà 2,4>1,5

nên \(\dfrac{12}{5}>\dfrac{3}{2}\)

b: \(\dfrac{2023}{2024}< \dfrac{2024}{2024}=1;\dfrac{2026}{2025}>\dfrac{2025}{2025}=1\)

Do đó: \(\dfrac{2023}{2024}< \dfrac{2026}{2025}\)

1. We enjoyed the city where we spent our vacation 

2. One of the elephants we saw at the zoo had only one tusk. 

3. The professor whose course I'm taking is excellent.

4. The man whose wallet was stolen called the police. 

5. A man brought in a small girl whose hand had been cut.

6. The Smiths, whose house had been destroyed in the explosion, were given rooms in the hotel. 

7. He introduced me to his students, most of whom were from abroad.

8. They gave me four very bad tyre, one of whichburst before I had drivem four miles.

9. A man who answered the phone said Tom was out.

10. Phuong Thao, whose music you like best, is a singer. 

a: Xét ΔBHA vuông tại H và ΔBAC vuông tại A có

\(\widehat{HBA}\) chung

Do đó: ΔBHA~ΔBAC
=>\(\dfrac{BH}{BA}=\dfrac{BA}{BC}\)

=>\(BH\cdot BC=BA^2\)

b: Xét ΔAHB vuông tại H có HD là đường cao

nên \(AD\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HE là đường cao

nên \(AE\cdot AC=AH^2\left(2\right)\)

Từ (1),(2) suy ra \(AD\cdot AB=AE\cdot AC\)

=>\(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)

Xét ΔADE vuông tại A và ΔACB vuông tại A có

\(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)

Do đó: ΔADE~ΔACB

đéo