\(d.2x-42=5.32.\Leftrightarrow2x-42=160.\Leftrightarrow2x=202\Leftrightarrow x=101.\)

\(e.140:\left(x-8\right)=7.\Leftrightarrow x-8=20.\Leftrightarrow x=28.\)

\(f.4\left(x+41\right)=400.\Leftrightarrow x+41=100\Leftrightarrow x=59.\)

Xin lỗi bạn hiền đang học lớp 5

1000 : [30 + (2^x - 6)] = 3² + 4²

1000 : [30 + (2^x - 6)] = 9 + 16

1000 : [30 + (2^x - 6)] = 25

30 + (2^x - 6) = 1000 : 25

30 + (2^x - 6) = 40

2^x - 6 = 40 - 30

2^x - 6 = 10

2^x = 10 + 6

2^x = 16

2^x = 2⁴

⇒ x = 4

(8,9 x 14,7 + 2,5 x 14,7) x (34 x 11 - 3400 x 0,1 - 34)

= (8,9 x 14,7 + 2,5 x 14,7) x (34 x 11 - 3400 : 10 - 34)

= [14,7 x (8,9 + 2,5)] x (34 x 11 - 34 x 10 - 34 x 1)

= (14,7 x 11,4) x [34 x (11 - 10 - 1)]

= (14,7 x 11,4) x (34 x 0)

= 0

 ( 14,7 x(8,9 + 2,5) ) x (34 x (11-100x0,1-1 )

=(14,7x11,4)x(34.(11-10-1))

=(14,7x11,4)x(34.0)

=(14,7x11,4)x0

=0

thanks ạ ☘

`2/[1xx5]+2/[5xx9]+2/[9xx13]+....+2/[93xx97]+2/[97xx101]`

`=1/2xx(4/[1xx5]+4/[5xx9]+4/[9xx13]+....+4/[93xx97]+4/[97xx101])`

`=1/2xx(1-1/5+1/5-1/9+1/9-1/13+...+1/93-1/97+1/97-1/101)`

`=1/2xx(1-1/101)`

`=1/2xx100/101`

`=50/101`

bn ghi rõ đc ko ạ mik ko hiểu lắm c.ơn bn

\(\dfrac{x^2+3x-4}{x-1}=\dfrac{x^2+4x-x-4}{\left(x-1\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{x-1}=x+4\)

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.4=16\\y^2=4.9=36\\z^2=16.16=16^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=6\\z=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-6\\z=-16\end{matrix}\right.\end{matrix}\right.\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)

Ta có: \(x^2+y^2+z^2=116\)

\(\Leftrightarrow4k^2+9k^2+16k^2=116\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=4k=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=4k=-8\end{matrix}\right.\)

Hhigh             

sửa đề : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

áp dụng t/c dãy t/s = nhau

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{58}{3+4+5}=\frac{58}{12}=\frac{29}{6}\)

\(\frac{x}{3}=\frac{29}{6}\Rightarrow x=\frac{29}{2}\)

\(\frac{y}{4}=\frac{29}{6}\Rightarrow y=\frac{58}{3}\)

\(\frac{z}{5}=\frac{29}{6}\Rightarrow z=\frac{145}{6}\)

vậy ...

chết rồi mình nhìn nhầm đề nhé : phải là :  3+4-5 = 58/2=29

từ đây bạn tìm x;y;z nhé!

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y-z}{3+4-5}=\frac{58}{2}=29\)

\(\Rightarrow\hept{\begin{cases}x=28.3=54\\y=28.4=112\\z=28.5=140\end{cases}}\)