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a) \(\dfrac{3,5}{15}=\dfrac{-2}{x}\)
\(\Rightarrow x=\dfrac{15.-2}{3,5}\)
\(\Rightarrow x=-8,57\)
b) \(2\left(3x-2\right)-3\left(x-2\right)-=-1\)
\(\Rightarrow6x-4-3x+6=-1\)
\(\Rightarrow6x-3x=-1+4-6\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-\dfrac{3}{3}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. Thay $y=x+1$ vào điều kiện ban đầu có:
$3x+5(x+1)=13$
$8x+5=13$
$8x=8$
$x=1$
$y=x+1=2$
b. Thay $x=y+5$ vô điều kiện đầu thì:
$2(y+5)-3y=4$
$-y+10=4$
$-y=-6$
$y=6$
$x=6+5=11$
c. Thay $y=x-2$ vô điều kiện đầu thì:
$-x+5(x-2)=-6$
$4x-10=-6$
$4x=10+(-6)=4$
$x=1$
$y=x-2=1-2=-1$
a) Ta có: \(\left\{{}\begin{matrix}3x+5y=13\\x+1=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\3x-3y=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8y=16\\x+1=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1=2-1=1\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}2x-3y=4\\x=y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\2x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=-6\\x=y+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=11\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}-x+5y=-6\\y=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+5y=-6\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=-4\\y=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=y+2=-1+2=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A(x)+B(x)=(x^3-3x^2+3x-1)+(2x^3+x^2-x+5)$
$=3x^3-2x^2+2x+4$
b.
$A(x)C(x)=(x^3-3x^2+3x-1)(x-2)=x(x^3-3x^2+3x-1)-2(x^3-3x^2+3x-1)$
$=(x^4-3x^3+3x^2-x)-(2x^3-6x^2+6x-2)$
$=x^4-5x^3+9x^2-7x+2$
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) \(x\)là giá trị tuyệt đối của 1 số nên \(x\ge0\)
\(\Rightarrow x.\left|x-4\right|=x\)
Với x = 0 :
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Với x > 0
TH1 : \(x< 4;\)ta có:
\(x.\left(4-x\right)=x\)
\(4x-x^2=x\)
\(x^2=4x-x=3x\Rightarrow x=3\)
TH2 : \(x\ge4;\)ta có:
\(x\left(x-4\right)=x\)
\(x^2-4x=x\)
\(\Rightarrow x^2=5x\)
\(\Rightarrow x=5\)
Vậy \(x\in\left\{0;3;4;5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
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Ta có: |x - 10| + 10 = x
=> |x - 10| = x - 10
=> x - 10 = x - 10
x - 10 = -(x - 10)
=> x - 10 = 0
=> x = 10 + 0
=> x = 10
\(2016^x=1\)
\(\Rightarrow x=0\)
tíc mình nha