Câu 1: She doesn't live in London because she doesn't know anyone there

->She'd.....live in London if she knew anyone there ......................................................................................................

Câu 2: I can't live in the country because I can't find a job there

->I'd...........live in the country if I could find a job there..............................................................................

Câu 3: I don't have a spare ticket. I can;t take you to the concert

->If I.........had a spare ticket, I could take you to the concert.................................

Câu 4: The schoolchildren sowed some seed, but they forgot to water them so they didn't go

-> If the children .....hadn't forgot to water the seeds, they would have grown................................

Câu 5: They don't understand a problem. They won't find a solution

-> If they...

Câu 6:He sit too much. He isn't fit

-> If he........didn't sit too much, he would be fit.....

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\(\dfrac{x}{y+z+1}\) = \(\dfrac{y}{x+z+2}\) = \(\dfrac{z}{x+y-3}\) = \(x+y+z\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{y+z+1}\)=\(\dfrac{y}{x+z+2}\)=\(\dfrac{z}{x+y-3}\)=\(\dfrac{x+y+z}{y+z+1+x+z+2+x+y-3}\)

\(x+y+z\) = \(\dfrac{x+y+z}{2.\left(x+y+z\right)}\) = \(\dfrac{1}{2}\) (1)

\(\dfrac{x}{y+z+1}\) = \(\dfrac{1}{2}\) ⇒ 2\(x\) = y+z+1 

⇒ 2\(x\) + \(x\) = \(x+y+z+1\) (2)

 Thay (1) vào (2) ta có: 2\(x\) + \(x\) = \(\dfrac{1}{2}\) + 1

                                      3\(x\)      = \(\dfrac{3}{2}\) ⇒ \(x=\dfrac{1}{2}\)

\(\dfrac{y}{x+z+2}\) = \(\dfrac{1}{2}\) ⇒ 2y = \(x+z+2\) ⇒ 2y+y = \(x+y+z+2\) (3)

Thay (1) vào (3) ta có: 2y + y = \(\dfrac{1}{2}\) + 2 

                                   3y = \(\dfrac{5}{2}\) ⇒ y = \(\dfrac{5}{6}\)

Thay \(x=\dfrac{1}{2};y=\dfrac{5}{6}\) vào (1) ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+z\) = \(\dfrac{1}{2}\)

                                                              \(\dfrac{5}{6}\) + z = 0 ⇒ z = - \(\dfrac{5}{6}\)

Kết luận: (\(x;y;z\)) = (\(\dfrac{1}{2}\); \(\dfrac{5}{6}\); - \(\dfrac{5}{6}\))

TH1: x + y + z $\ne$ 0

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

$\frac{x}{y+z+1}$ = $\frac{y}{x+z+2}$ = $\frac{z}{x+y-3}$ = $\frac{x+y+z}{y+z+1+x+z+2+x+y-3}$ 

              = $\frac{x+y+z}{x+y+z+x+y+z}$ = $\frac{x+y+z}{2(x+y+z)}$ = $\frac{1}{2}$ 

⇒ x + y + z = $\frac{1}{2}$

⇒ x + y       = $\frac{1}{2}$ - z

    x + z        = $\frac{1}{2}$ - y

    y + z        = $\frac{1}{2}$ - x

Thay y + z + 1 = $\frac{1}{2}$ - x + 1

⇒ $\frac{x}{\frac{1}{2}-x+1}$ = $\frac{1}{2}$

⇒ 2x = $\frac{1}{2}$ - x + 1

⇒ 2x + x = $\frac{1}{2}$ + 1

⇒  3x   =  $\frac{3}{2}$

⇒   x    = $\frac{1}{2}$

Thay x + z + 2 = $\frac{1}{2}$ - y + 2

⇒ $\frac{y}{\frac{1}{2}-y+2}$ = $\frac{1}{2}$

⇒ 2y = $\frac{1}{2}$ - y + 2

⇒ 2y + y = $\frac{1}{2}$ + 2

⇒   3y  = $\frac{5}{2}$

⇒     y   = $\frac{5}{6}$

Thay x + y - 3 = $\frac{1}{2}$ - z - 3

⇒ $\frac{z}{\frac{1}{2}-z-3}=$\frac{1}{2}$

⇒ 2z = $\frac{1}{2}$ - z - 3

⇒ 2z + z = $\frac{1}{2}$ - 3

⇒  3z  = $\frac{-5}{2}$

⇒   z   = $\frac{-5}{6}$

TH2: x + y + z = 0

⇒ $\frac{x}{y+z+1}$ = $\frac{y}{x+z+2}$ = $\frac{z}{x+y-3}$ = 0

⇒ x = y = z = 0

https://olm.vn/cau-hoi/tim-tat-ca-cac-so-xyz-biet-dfracxyz1dfracyxz2dfraczxy-3xyz-giair-chi-tiet-ho-e-vs-a.8297156371934

\(6,\\ a,P=9\left(x^2-2\cdot\dfrac{1}{9}x+\dfrac{1}{81}\right)+\dfrac{26}{9}=9\left(x-\dfrac{1}{9}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\\ P_{min}=\dfrac{26}{9}\Leftrightarrow x-\dfrac{1}{9}=0\Leftrightarrow x=\dfrac{1}{9}\\ b,Q=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\\ Q_{min}=\dfrac{1}{4}\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\\ c,R=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\\ R_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

\(\Leftrightarrow\dfrac{90}{x}-\dfrac{90}{x+5}=\dfrac{1}{5}\)

\(\Leftrightarrow x\left(x+5\right)=450x+2250-450x\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=45\\x=-50\end{matrix}\right.\)

ĐKXĐ: \(x\ne0;x\ne-5\)

\(\dfrac{180}{x}-\dfrac{180}{x+5}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{180.5.\left(x+5\right)}{5x\left(x+5\right)}-\dfrac{180.5.x}{5x\left(x+5\right)}=\dfrac{2.x.\left(x+5\right)}{5x\left(x+5\right)}\)

\(\Leftrightarrow900x+4500-900x=2x^2+10x\)

\(\Leftrightarrow2x^2+10x-4500=0\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow x^2+50x-45x-2250=0\)

\(\Leftrightarrow x\left(x+50\right)-45\left(x+50\right)=0\)

\(\Leftrightarrow\left(x+50\right)\left(x-45\right)=0\)

=>   x+50=0         hoặc                 x-45=0

         x=-50                                    x=45

Vậy....

c) Ta đã biết DA = DE (chứng minh trên) (1)

Trong tam giác EDC vuông tại E có DC đối diện đỉnh E

Suy ra DC là cạnh lớn nhất trong tam giác EDC

Hay DC > DE (2)

Từ (1) và (2) suy ra DC > AD hay AD < DC

giải hộ mình vớiii