6 - 7x = -48
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`4)x^2-5x+6`
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)=(x-2)(x-3)`
`5)x^2+7x+10`
`=x^2+5x+2x+10`
`=x(x+5)+2(x+5)=(x+5)(x+2)`
`6)x+7\sqrt{x}+10` `ĐK: x >= 0`
`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`
`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`
`7)3x^4+7x^2+4`
`=3x^4+3x^2+4x^2+4`
`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`
`8)x^2-x-2`
`=x^2-2x+x-2`
`=x(x-2)+(x-2)=(x-2)(x+1)`
`9)x^6-x^3-2`
`=x^6+x^3-2x^3-2`
`=x^3(x^3+1)-2(x^3+1)`
`=(x^3+1)(x^3-2)`.
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a) y - 6 : 2 - (48 - 24 x 2 : 6 - 3) = 0
y - 3 - (48 - 48 : 6 - 3) = 0
y - 3 - (48 - 8 - 3) = 0
y - 3 - 37 = 0
y - ( 3+37) = 0
y - 40 =0
y =0+40
Y =40
b) ( 7x13 - 8x13) : ( 9 2/3 -y) =39
(7+8)x13 : (29/3 - y) =39
15 x 13 : (29/3-y) =39
195 : (29/3 - y) =39
29/3 - y =195 : 39
29/3 - y = 5
y = 29/3 - 5
y = 14/3
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Lời giải:
$f(x)=5x^2.0,4623x^{87}+7x^{46}$
$=2,3115x^{89}+7x^{46}$
$=x^{46}(2,3115x^{43}+7)=0$
$\Leftrightarrow x^{46}=0$ hoặc $2,3115x^{43}+7=0$
$\Rightarrow x=0$ hoặc $x=\sqrt[43]{\frac{-7}{2,3115}}$
Đây chính là 2 nghiệm của đa thức.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)y-6:2-\left(48-24x2:6-3\right)=0\)\(0\)
\(y-3-\left(48-48:6-3\right)=0\)
\(y-3-\left(48-8-3\right)=0\)
\(y-3-\left(40-3\right)=0\)
\(y-3-37=0\)
\(y-\left(3+37\right)=0\)
\(y-40=0\)
\(y=0+40\)
\(y=40\)
\(b)\left(7x13+8x13\right):\left(9\frac{2}{3}-y\right)=39\)
\(\left(91+104\right):\left(\frac{29}{3}-y\right)=39\)
\(195:\left(\frac{29}{3}-y\right)=39\)
\(\frac{29}{3}-y=195:39\)
\(\frac{29}{3}-y=5\)
\(y=\frac{29}{3}-5\)
\(y=\frac{14}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)
\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)
\(=0\)
Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
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6.
a) \(7x-35=-\left(-2x+5\right)\\ \Leftrightarrow7x-35=2x-5\\ \Leftrightarrow7x-2x=35-5\\ \Leftrightarrow5x=30\\ \Leftrightarrow x=6\)Vậy pt có tập nghiệm \(S=\left\{6\right\}\)
b) \(4x-x-18=3\left(x-6\right)\\ \Leftrightarrow3x-18=3x-6\\ \Leftrightarrow3x-3x=18-6\\ \Leftrightarrow0=12\left(\text{vô lí}\right)\)Vậy pt có tập nghiệm \(S=\varnothing\)
c) \(x-6=8+x\\ \Leftrightarrow x-x=8+6\\ \Leftrightarrow0=14\left(\text{vô lí}\right)\)Vậy pt có tập nghiệm \(S=\varnothing\)
d) \(48-5x=39-\left(-2x\right)\\ \Leftrightarrow48-5x=39+2x\\ \Leftrightarrow48-39=5x+2x\\ \Leftrightarrow7x=9\\ \Leftrightarrow x=\frac{9}{7}\)Vậy pt có tập nghiệm \(S=\left\{\frac{9}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)
\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)
=2x^2-3x+4
b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)
\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)
c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
6-7x=-48
7x=-48+6
7x=-42
x=-42:7
x=-6
k nha
Trl:
\(6-7x=-48\)
\(\Rightarrow7x=-48+6\)
\(\Rightarrow7x=-42\)
\(\Rightarrow x=-42:7\)
\(\Rightarrow x=-6\)