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*Tính M(x) - N(x)
M(x) = -x3 + 2,5x2 - 0,5x - 1
N(x) = -x3 + 2,5x2 + 2x - 6
------------------------------------
M(x) - N(x) = -2,5x + 5
=> M(x) - N(x) = A(x) = -2,5x + 5
Để đa thức A(x) có nghiệm => -2,5x + 5 = 0
=> -2,5x = -5
=> 2,5x = 5
=> x = 2
Tính M(x) + N(x)
M(x) = -x3 + 2,5x2 - 0,5x - 1
N(x) = -x3 + 2,5x2 + 2x - 6
---------------------------------------------
M(x) + N(x) = -2x3 + 5x2 + 1,5x - 7
=> M(x) + N(x) = B(x) = -2x3 + 5x2 + 1,5x - 7
Bậc của đa thức B(x) là 3
P/S : Cái dấu chấm đó là nhân hay phẩy?
![](https://rs.olm.vn/images/avt/0.png?1311)
mk biết có mỗi cách thôi, làm xằng vậy =))
Ta có :
\(5x=3y\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt :
\(\dfrac{x}{3}=\dfrac{y}{5}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)\(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xy=375\) ta có :
\(3k.5k=375\)
\(\Leftrightarrow15k^2=375\)
\(\Leftrightarrow\left[{}\begin{matrix}k=25\\k=-25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
+) \(k=5\Leftrightarrow\left[{}\begin{matrix}x=3.5=15\\y=5.5=25\end{matrix}\right.\)
+) \(k=-5\Leftrightarrow\left[{}\begin{matrix}x=3.\left(-5\right)=-15\\y=5\left(-5\right)=-25\end{matrix}\right.\)
Vậy ..
1 Cách nhé
\(5x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt : \(\dfrac{x}{3}=\dfrac{y}{5}=k\)
Ta có : \(\dfrac{x}{3}=k\Leftrightarrow x=3k\)
\(\dfrac{y}{5}=k\Leftrightarrow y=5k\)
Mà : \(3k.5k=375\)
\(\Leftrightarrow15k^2=375\)
\(\Leftrightarrow k^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
Khi \(k=5\), thì : \(\left\{{}\begin{matrix}x=15\\y=25\end{matrix}\right.\)
Khi \(k=-5,\) thì : \(\left\{{}\begin{matrix}x=-15\\y=-25\end{matrix}\right.\)
Vậy ..........
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b. \(\left|2,5x-x\right|=1,3\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=1,3\\2,5x-x=-1,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{15}\\x=-\dfrac{13}{15}\end{matrix}\right.\)
d. \(\left(3x-2\right)^3=-27\)
\(\Leftrightarrow\left(3x-2\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Lời giải:
a.
$|2,5x-x|=1,3$
$|1,5x|=1,3$
$\Rightarrow 1,5x=\pm 1,3$
$\Rightarrow x=\pm \frac{13}{15}$
b.
$(3x-2)^3=(-27)=(-3)^3$
$\Rightarrow 3x-2=-3$
$\Rightarrow 3x=-1$
$\Rightarrow x=\frac{-1}{3}$
![](https://rs.olm.vn/images/avt/0.png?1311)
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a)\(\left(1-x\right)^3=216\)
\(\Rightarrow1-x=6\)
\(\Rightarrow x=-5\)
b)\(3^{x+1}-3^x=162\)
\(\Rightarrow3^x\left(3-1\right)=162\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
c)\(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Tìm x biết :
5x+1-2.5x= 375
5x.5-2.5x= 375
5x=375:3
5x=125
x=3
\(5^{x+1}-2.5^x=375\)
\(5^x.5-2.5^x=375\)
\(5^x.\left(5-2\right)=375\)
\(5^x.3=375\)
\(5^x=375:3\)
\(5^x=125\)
\(5^x=5^3\)
VẬY => x=3