Áp dụng tính chất của dãy tỉ số bằng nhau 

\(\Rightarrow\frac{2+3y}{13}=\frac{2+6y}{17}=\frac{2+9y}{18}=\)\(\frac{\left(2+9y\right)-\left(2+3y+2+6y\right)}{18-\left(13+17\right)}=\frac{-2}{-2}\)\(=1\)

\(\Rightarrow2+3y=13\Rightarrow3y=11\Rightarrow y=\frac{11}{3}\)

Vậy \(y=\frac{11}{3}\)

\(\frac{2+3y}{13}\)= \(\frac{2+6y}{17}\)= \(\frac{2+9y}{18}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có \(\frac{2+3y}{13}\) = :\(\frac{2+6y}{17}\) = \(\frac{2+9y}{18}\) = \(\frac{2+3y+2+6y-2-9y}{13+17-18}\)= \(\frac{2}{12}\)= \(\frac{1}{6}\)
\(\Rightarrow\frac{2+3y}{13}\)= \(\frac{1}{6}\)\(\Rightarrow2+3y=\frac{1}{6}\)x 13 = \(\frac{13}{6}\)\(\Rightarrow3y=\frac{13}{6}\)- 2 \(\Rightarrow3y=\frac{1}{6}\)\(\Rightarrow y=\frac{1}{18}\)
Chúc bạn học tốt!

Giúp mình với ạ,cảm ơn mọi người

b: Ta có: \(B=x^2+4x+9y^2-6y-1\)

\(=x^2+4x+4+9y^2-6y+1-6\)

\(=\left(x+2\right)^2+\left(3y-1\right)^2-6\ge-6\forall x,y\)

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)

Bài 2:

a: \(\Leftrightarrow4x^2-14x+10x-35-\left(4x+3\right)^2=16\)

\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)

\(\Leftrightarrow-12x^2-28x-60=0\)

\(\Leftrightarrow3x^2+7x+15=0\)

\(\text{Δ}=7^2-4\cdot3\cdot15=-131< 0\)

Do đó: Phương trình vô nghiệm

b: Ta có: \(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1\right)^2=22\)

\(\Leftrightarrow64x^4-9-64x^4+16x^2-1=22\)

\(\Leftrightarrow16x^2=32\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

c: Ta có: \(49x^2+14x+1=0\)

=>\(\left(7x+1\right)^2=0\)

hay x=-1/7

`y=2/3x`

`=>3y=2x`

`=>8x=12y`

Mặt khác:`4z=3y`

`=>z=3/4y`

`=>5z=15/4y`

Thay `8x=12y,5z=15/4y` vào `8x+9y+5z=1980`

`=>15/4y+9y+12y=1980`

`=>21y+15/4y=1980`

`=>99/4y=1980`

`=>1/4y=20`

`=>y=80`

`=>x=3/2y=120,z=3/4y=60`

Vậy `(x,y,z)=(120,80,60)`

Ta có: 4z=3y

nên \(4z=3\cdot\dfrac{2}{3}x=x\)

hay \(z=\dfrac{1}{4}x\)

Ta có: 8x+9y+5z=1980

\(\Leftrightarrow8x+9\cdot\dfrac{2}{3}x+5\cdot\dfrac{1}{4}x=1980\)

\(\Leftrightarrow x\cdot\dfrac{61}{4}=1980\)

hay \(x=\dfrac{7920}{61}\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{3}x=\dfrac{2}{3}\cdot\dfrac{7920}{61}=\dfrac{5280}{61}\\4z=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\4z=\dfrac{15840}{61}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5280}{61}\\z=\dfrac{3960}{61}\end{matrix}\right.\)

Vậy: \(\left(x,y,z\right)=\left(\dfrac{7920}{61};\dfrac{5280}{61};\dfrac{3960}{61}\right)\)

Giải:

Ta có:

\(y=\dfrac{2}{3}x;4z=3y\) 

\(\Rightarrow x=\dfrac{3}{2}y;z=\dfrac{3}{4}y\)

Theo đề bài ta có:

\(8x+9y+5z=1980\) 

\(\Rightarrow8.\dfrac{3}{2}y+9y+5.\dfrac{3}{4}y=1980\) 

\(\Rightarrow12y+9y+\dfrac{15}{4}y=1980\)  

\(\Rightarrow y.\left(12+9+\dfrac{15}{4}\right)=1980\) 

\(\Rightarrow y.\dfrac{99}{4}=1980\) 

\(\Rightarrow y=1980:\dfrac{99}{4}\) 

\(\Rightarrow y=80\) 

\(\Rightarrow x=\dfrac{3}{2}.80=120\) 

\(\Rightarrow z=\dfrac{3}{4}.80=60\) 

Vậy \(\left(x;y;z\right)=\left(120;80;60\right)\) 

Bài 2:

a: \(\Leftrightarrow4x^2-14x+10x-35-\left(4x+3\right)^2=16\)

\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)

\(\Leftrightarrow-12x^2-28x-60=0\)

\(\Leftrightarrow3x^2+7x+15=0\)

\(\text{Δ}=7^2-4\cdot3\cdot15=-131< 0\)

Do đó: Phương trình vô nghiệm

b: Ta có: \(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1\right)^2=22\)

\(\Leftrightarrow64x^4-9-64x^4+16x^2-1=22\)

\(\Leftrightarrow16x^2=32\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

c: Ta có: \(49x^2+14x+1=0\)

=>\(\left(7x+1\right)^2=0\)

hay x=-1/7

Mình ko bít ?

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

\(=\dfrac{\left(2x-3y\right)^3}{\left(2x-3y\right)^2}=2x-3y\)

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3