`y xx1/2+y xx1/4+y xx1/8=22/24`

`=>y xx (1/2+1/4+1/8)=11/12`

`=> y xx (4/8+2/8+1/8)=11/12`

`=> y xx 7/8=11/12`

`=>y=11/12:7/8`

`=>y=11/12xx8/7`

`=>y=22/21`

`y xx 1/2 + y xx 1/4 + y xx 1/8 = 22/24`

`=> y xx (1/2 + 1/4 + 1/8) = 11/12`

`=> y xx ( 4/8 + 2/8+ 1/8) = 11/12`

`=> y xx 7/8 = 11/12`

`=> y= 11/12 : 7/8`

`=> y=11/12 xx 8/7`

`=> y=22/21`

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

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

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)

\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)

\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)

\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)

\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)

.... mấy ý còn lại bn tự lm nhé, tương tự thhooi

1) \(-\left(y+6\right)^2=-y^2-12y-36\)

2) \(-\left(4-y\right)^2=-y^2+8y-16\)

3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)

4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)

5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)

6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)

7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)

8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)

\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\) (sửa đề)

\(\Rightarrow3+2^{x-1}=24-\left[16-\left(4-1\right)\right]\)

\(\Rightarrow3+2^{x-1}=24-\left(16-3\right)\)

\(\Rightarrow3+2^{x-1}=24-13\)

\(\Rightarrow3+2^{x-1}=11\)

\(\Rightarrow2^{x-1}=11-3\)

\(\Rightarrow2^{x-1}=8\)

\(\Rightarrow2^{x-1}=2^3\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x=3+1=4\)

Vậy \(x=4.\)

#\(Toru\)

a) Ta có:

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

Áp dụng tính chất dãy tỉ số bằng nhau ta được

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

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

\(=\dfrac{22+2}{2}\)

\(=\dfrac{24}{2}=12\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{3}=12\\\dfrac{y+2}{2}=12\\\dfrac{z+3}{1}=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\y+2=24\\z+3=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=36-1=35\\y=24-2=22\\z=12-3=9\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)