2x3^x+1-3^x-1=3^26+8x27^8
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2x3x=10x312+8x274
2x3x=10x312+8x312
2x3x=(10+8)x312
2x3x=18x312
2x3x=2x32x312
2x3x=2x314
=>3x=314 =>x=14
2x3x=10x312+8x274
2x3x=10x531441+8x531441
2x3x=(10+8)x531441
2x3x=18x531441
2x3x=2x9x531441
3x=9x531441
3x=4782969
3x=314
=>x=14
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
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2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
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3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
a)(9x+2)x3=60
9x+2 =60:3
9x+2 =20
9x =20-2
9x=18
x=18:9
x=2
\(C,\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{2}.\dfrac{4}{4}=\dfrac{1}{2}.1=\dfrac{1}{2}\)
\(D,\dfrac{11}{3}.\dfrac{26}{7}-\dfrac{26}{7}.\dfrac{8}{3}=\dfrac{26}{7}.\left(\dfrac{11}{3}-\dfrac{8}{3}\right)\)
\(=\dfrac{26}{7}.\dfrac{3}{3}=\dfrac{26}{7}.1=\dfrac{26}{7}\)
c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)
\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)
\(=5x^3+14x^2+12x+8\)
d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)
\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)
\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)
\(=5x^2+4x+4\)