4(x+41)=400 //mn tính giúp mik với ạ///^^
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\(d.2x-42=5.32.\Leftrightarrow2x-42=160.\Leftrightarrow2x=202\Leftrightarrow x=101.\)
\(e.140:\left(x-8\right)=7.\Leftrightarrow x-8=20.\Leftrightarrow x=28.\)
\(f.4\left(x+41\right)=400.\Leftrightarrow x+41=100\Leftrightarrow x=59.\)
bạn có thể làm rõ ràng hơn đc hok. Tại mình nhìn nó hơi rối
(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
`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`
\(\dfrac{x^2+3x-4}{x-1}=\dfrac{x^2+4x-x-4}{\left(x-1\right)}=\dfrac{\left(x+4\right)\left(x-1\right)}{x-1}=x+4\)
\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4.4=16\\y^2=4.9=36\\z^2=16.16=16^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=6\\z=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-6\\z=-16\end{matrix}\right.\end{matrix}\right.\)
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 ...
\(\Rightarrow x+41=400:4=100\\ \Rightarrow x=100-41=59\)
x+41=400:4
x =100-41
⇔x =59