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(x−3)(x2+3x+9)−(3x−17)=x3−12(x−3)(x2+3x+9)−(3x−17)=x3−12
⇒x(x2+3x+9)−3(x2+3x+9)−3x+17=x3−12⇒x(x2+3x+9)−3(x2+3x+9)−3x+17=x3−12
⇒x3+3x2+9x−3x2−9x−27−3x+17=x3−12⇒x3+3x2+9x−3x2−9x−27−3x+17=x3−12
⇒x3+(3x2−3x2)+(9x−9x)−3x−10=x3+12⇒x3+(3x2−3x2)+(9x−9x)−3x−10=x3+12
⇒x3−3x−10=x3+12⇒x3−3x−10=x3+12
⇒x3−3x−10−12=x3⇒x3−3x−10−12=x3
⇒x3−3x−22=x3⇒x3−3x−22=x3
⇒3x−22=0⇒3x−22=0
⇒3x=22⇒x=223
(x−3)(x^2+3x+9)−(3x−17)=x^3−12
⇔x^3−27−3x+17=x^3−12
⇔−10−3x=−12
⇔3x=2
⇔x=2/3
Vậy...
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`@` `\text {Ans}`
`\downarrow`
Thực hiện phép tính ;-;?
\((x+3) (x^2-3x+9) + (x-3) ( x^2+3x+9 )\)
`= x(x^2 - 3x + 9) + 3(x^2 - 3x + 9) + x(x^2 + 3x + 9) - 3(x^2 + 3x + 9)`
`= x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 + x^3 + 3x^2 + 9x - 3x^2 - 9x - 27`
`= (x^3 + x^3) + (-3x^2 + 3x^2 + 3x^2 - 3x^2) + (9x - 9x + 9x - 9x) + (27 - 27)`
`= 2x^3`
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a. M(x) + N(x) = 3x3 - 3x + x2 + 5 + 2x2 - x + 3x3 + 9
= (3x3 + 3x3) + ( x2 + 2x2 ) + ( -3x - x ) + (5 + 9)
= 6x3 + 3x2 - 4x + 14
b. M(x) + N(x) - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - P(x) = 6x3 + 3x2 + 2x
=> 6x3 + 3x2 - 4x + 14 - ( 6x3 + 3x2 + 2x) = P(x)
=> 6x3 + 3x2 - 4x + 14 - 6x3 - 3x2 - 2x = P(x)
=> (6x3 - 6x3 ) + (3x2 - 3x2 ) + (-4x - 2x ) + 14 = P(x)
=> -6x + 14 = P(x)
Ta có : -6x + 14 = 0
=> -6x = -14
=> x = 7/3
=> Đa thức P(x) = -6x + 14 có nghiệm là 7/3
=>
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* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
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a,\(4\left(5x-3\right)-3\left(2x+1\right)=9\)
\(\Leftrightarrow20-12-6x-3=9\)
\(\Leftrightarrow14x=9+12+3\)
\(\Leftrightarrow14x=24\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy.....
b, |x-9|=2x+5
* Với x ≥ 9 thì |x – 9| = x – 9 ta có PT: x – 9 = 2x + 5
\(\Leftrightarrow\) x = - 14 ( loại)
*Với x < 9 thì |x – 9| = 9 – x ta có PT: 9 – x = 2x + 5
\(\Leftrightarrow\) x = 4/3(thỏa mãn)
=>Vậy phương trình có tập nghiệm là S = \(\left\{\dfrac{4}{3}\right\}\)
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\(\frac{1}{3}x.\left(x-2\right)-3x.\left(x+\frac{1}{3}\right)=-9\)
\(\Rightarrow\frac{1}{3}x^2-\frac{2}{3}x-\left(3x^2+x\right)=-9\)
\(\Rightarrow\frac{1}{3}x^2-\frac{2}{3}x-3x^2-x=-9\)
\(\Rightarrow\left(\frac{1}{3}x^2-3x^2\right)-\left(\frac{2}{3}x+x\right)=-9\)
\(\Rightarrow-\frac{8}{3}x^2-\frac{5}{3}x=-9\)
\(\Rightarrow-\frac{8}{3}x^2-\frac{5}{3}x+9=0\)
Chịu.