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\(x+\dfrac{2}{4}=\dfrac{3}{6}\\ x+\dfrac{1}{2}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{1}{2}\\ x=0\)
\(x+\)\(2/4\)\(=\)\(\dfrac{3}{6}\)
\(_x\)\(\) \(=\)\(3/6\)\(-\)\(\dfrac{2}{4}\)
x =0
Sửa đề: 12x37x4+8x41x6+3x22x16
=\(48\cdot37+48\cdot41+48\cdot22\)
\(=48\left(37+41+22\right)=48\cdot100=4800\)
12 x 37 x 4 + 8 x 41 x 6 + 3 x 22 x 16
= (12 x 4) x 37 + (8 x 6) x 41 + (3 x 16) x 22
= 48 x 37 + 48 x 41 + 48 x 22
= 48 x (37 + 41 + 22)
= 48 x (78 + 22)
= 48 x 100
= 4800
`2,4 : (2-x) - 0,32 . 4,5 = 1,56`
`=> 2,4 : (2-x) - 1,44 = 1,56`
`=> 2,4 : (2-x) = 1,56 + 1,44`
`=> 2,4 : (2-x) = 3`
`=> 2 - x = 2,4 : 3`
`=> 2 - x = 0,8`
`=> x = 2 - 0.8`
`=> x = 1,2`
Vậy `x = 1,2`
\(2,4:\left(2-x\right)-0,32\cdot4,5=1,56\)
=>\(2,4:\left(2-x\right)=1,56+0,32\cdot4,5=3\)
=>2-x=2,4:3=0,8
=>x=2-0,8=1,2
\(x\cdot9,63-5,63\cdot x=15,6\)
=>\(x\left(9,63-5,63\right)=15,6\)
=>4x=15,6
=>x=15,6:4=3,9
\(\dfrac{5}{x}-\dfrac{2}{y}=\dfrac{3}{2}\)
=>\(\dfrac{5x-2y}{xy}=\dfrac{3}{2}\)
=>2(5x-2y)=3xy
=>10x-4y-3xy=0
=>10x-3xy-4y=0
=>x(10-3y)-4y=0
=>\(-3x\left(y-\dfrac{10}{3}\right)-4y+\dfrac{40}{3}=0\)
=>\(-3x\left(y-\dfrac{10}{3}\right)-4\left(y-\dfrac{10}{3}\right)=0\)
=>\(\left(-3x-4\right)\left(y-\dfrac{10}{3}\right)=0\)
=>\(\left\{{}\begin{matrix}-3x-4=0\\y-\dfrac{10}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\y=\dfrac{10}{3}\end{matrix}\right.\)
Lời giải:
Chiều rộng thửa ruộng:
$60\times 40:100=24$ (m)
Diện tích thửa ruộng:
$60\times 24=1440$ (m2)
Thửa ruộng đó thu hoạch được số kg khoai tây là:
$1440:100\times 240=3456$ (kg)
Số tiền thu được khi bán khoai tây là:
$3456\times 25000=86400000$ (đồng)
\(|x^2|x+\dfrac{3}{4}||=x^2\)
=>\(x^2\cdot\left|x+\dfrac{3}{4}\right|=x^2\)
=>\(\left|x+\dfrac{3}{4}\right|=1\)
=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=1\\x+\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
|\(x^2\).|\(x+\dfrac{3}{4}\)| |= \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| = \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| - \(x^2\) = 0
\(x^2\).(|\(x+\dfrac{3}{4}\)| - 1) = 0
\(\left[{}\begin{matrix}x=0\\\left|x+\dfrac{3}{4}\right|=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x+\dfrac{3}{4}=-1\\x+\dfrac{3}{4}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{7}{4}\); 0; \(\dfrac{1}{4}\)}
\(\dfrac{1}{x^2\left(y-z\right)}=-\dfrac{3}{5}\Rightarrow x^2=-\dfrac{5}{3\left(y-z\right)}\)
\(\dfrac{1}{y^2\left(z-x\right)}=\dfrac{1}{3}\Rightarrow y^2=\dfrac{3}{\left(z-x\right)}\)
\(\dfrac{1}{z^2\left(x-y\right)}=3\Rightarrow z^2=\dfrac{1}{3\left(x-y\right)}\)
\(A=x^2.y^2.z^2=-\dfrac{5}{3\left(y-z\right)}.\dfrac{3}{z-x}.\dfrac{1}{3\left(x-y\right)}=\)
\(=-\dfrac{5}{3}.\dfrac{1}{\left(y-z\right)\left(z-x\right)\left(x-y\right)}=\)
1 He isn't old enough to join the club
2 Their house will be repained next year
3 Lan suggested going to the movies
4 It takes me three hours to drive from here to HCM city
5 You'd better take a rest
6 Despite being old, he can ride a bike to work
7 I apologise to you for breaking the vase
8 The sea is too rough for us to swim