gấp lắm ạ
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ND
2
giúp em với ạ e, đang cần gấp lắm ạ
H
2 tháng 7 2023
\(b,\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\\ =\left(\dfrac{1}{2}+\dfrac{5}{18}\right)+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}\\ =\left(\dfrac{9}{18}+\dfrac{5}{18}\right)+\dfrac{19}{19}-\dfrac{4}{9}\\ =\dfrac{14}{18}+1-\dfrac{4}{9}\\ =\dfrac{7}{9}+1-\dfrac{4}{9}\\ =\left(\dfrac{7}{9}-\dfrac{4}{9}\right)+1\\ =\dfrac{3}{9}+1\\ =\dfrac{1}{3}+1\\ =\dfrac{4}{3}\)
\(c,\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\\ =\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\left(\dfrac{2}{5}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-\dfrac{23}{23}+\left(\dfrac{6}{15}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-1+\dfrac{13}{15}+\dfrac{2}{3}\\ =-\dfrac{15}{15}+\dfrac{13}{15}+\dfrac{10}{15}\\ =\dfrac{8}{15}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\\ =\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}.\dfrac{-7}{11}\\ =-\dfrac{35}{77}\\ =-\dfrac{5}{11}\)
\(f,\dfrac{2}{11}.\dfrac{-5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+1\dfrac{3}{4}\\ =-\dfrac{2}{11}.\dfrac{5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{5}{4}.\left(-\dfrac{2}{11}+\dfrac{-9}{11}\right)+\dfrac{7}{4}\\ =\dfrac{5}{4}.1+\dfrac{7}{4}\\ =\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{12}{4}\\ =3\)
\(h,\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{5}\cdot\dfrac{9}{4}+3\dfrac{2}{13}\\ =\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{4}\cdot\dfrac{9}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\left(\dfrac{29}{5}-\dfrac{9}{5}\right)+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\dfrac{20}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}.4+\dfrac{41}{13}\\ =\dfrac{28}{4}+\dfrac{41}{13}\\ =7+\dfrac{41}{13}\\ =\dfrac{132}{13}\)
L
9 tháng 5 2022
a) P (x) = M (x) + N (x) = -2x2 + 3 - 4x3 + 5x + 4x3 + 10 + 2x2 - 3x
= ( -2x2 + 2x2 ) + ( 3 + 10 ) + ( -4x3 + 4x3 ) + ( 5x - 3x )
= 13 + 2x
Ta có P (x) = 13 + 2x
Đặt 13 + 2x = 0
=> 2x = - 13
=> x = \(\dfrac{-13}{2}\)
Vậy nghiệm của đa thức P (x) là \(\dfrac{-13}{2}\)
b) Q (x) = M (x) - N (x)
=> Q (x) = -2x2 + 3 - 4x3 + 5x - ( 4x3 + 10 + 2x2 - 3x)
= -2x2 + 3 - 4x3 + 5x - 4x3 - 10 - 2x2 + 3x
= ( -2x2 - 2x2 ) + ( 3 - 10 ) + ( -4x3 - 4x3 ) + ( 5x + 3x )
= -4x2 - 7 - 8x3 + 8x
PT
9 tháng 5 2022
a) P (x) = M (x) + N (x) = -2x2 + 3 - 4x3 + 5x + 4x3 + 10 + 2x2 - 3x
= ( -2x2 + 2x2 ) + ( 3 + 10 ) + ( -4x3 + 4x3 ) + ( 5x - 3x )
= 13 + 2x
Ta có P (x) = 13 + 2x
Đặt 13 + 2x = 0
=> 2x = - 13
=> x = −132−132
Vậy nghiệm của đa thức P (x) là −132−132
b) Q (x) = M (x) - N (x)
=> Q (x) = -2x2 + 3 - 4x3 + 5x - ( 4x3 + 10 + 2x2 - 3x)
= -2x2 + 3 - 4x3 + 5x - 4x3 - 10 - 2x2 + 3x
= ( -2x2 - 2x2 ) + ( 3 - 10 ) + ( -4x3 - 4x3 ) + ( 5x + 3x )
= -4x2 - 7 - 8x3 + 8x
gấp lắm ạ chỉ em với :<
giúp em với ạ em đang cần gấp lắm ạ em cảm ơn trước
TM
3 tháng 7 2023
(a) \(A=\dfrac{3}{x-2}\in Z\)
\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;0;2;4\right\}.\)
(b) \(B=-\dfrac{11}{2x-3}\in Z\)
\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;1;2;7\right\}.\)
(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)
\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)
(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)
\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)
4 tháng 12 2021
a: Xét ΔABH vuông tại H và ΔMBH vuông tại H có
HB chung
HA=HM
Do đó: ΔABH=ΔMBH
a: Ta có: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|7-x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)