tìm x biết
2-x/-3 = 27/x-2
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a. x × 3 = 27
x = 27 : 3
x = 9
b. 4 × x = 20
x = 20 : 4
x = 5
c. 10 + x : 2 = 20
x : 2 = 20 – 10
x : 2 = 10
x = 10 × 2
x = 20
d. x × 3 = 27 + 3
x × 3 = 30
x = 30 : 3
x = 10
e. 27 : x = 789 – 780
27 : x = 9
x = 27 : 9
x = 3
(x + 3)(x2 - 3x + 9) - x(x - 2)2 = 27
\(\Leftrightarrow\) x3 + 27 - x( x2 - 4x + 4) = 27
\(\Leftrightarrow\) x3 + 27 - x3 + 4x2 - 4x - 27 = 0
\(\Leftrightarrow\) 4x2 - 4x = 0
\(\Leftrightarrow\) 4x ( x - 1) = 0
khi 4x = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) x = 0 \(\Leftrightarrow\) x = 1
Chúc bạn học tốt
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.2^2=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+4x=27\\ 4x+27=27\\ 4x=0\\ x=0\)
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)
`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
27:(x-3/2)^3=(x-3/2):3
Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)
x^3 = 27
x^3 = 3^3
=> x = 3
(2x-1)^3 = 8
(2x-1)^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3:2
=> ko có x phù hợp.
(x-2)^2 = 16
(x-2)^2 = 4^2
x-2 = 4
x = 4+2
x = 6
Chúc bn học tốt!
x^3 = 27
x^3 = 3^3
Vậy x = 3
Đề bài 2 hình như sai bạn ạ
( x - 2 )^2 = 16
( x- 2 )^2 = 4^2
x - 2 = 4
x = 4 + 2
x = 6
Vậy x = 6
b) \(x^2\) : 3 = 27 \(x^2\) = 27 x 3 x = 81 TH1: x = 9 TH2: x = (-9) Vậy x ϵ (-9, 9)
Ta có\(\frac{2-x}{-3}=\frac{27}{x-2}\)
(2 - x)(x - 2) = 27.(-3)
=> -(x - 2)2 = - 81
=> (x - 2)2 = 81
=> \(\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)
\(\frac{2-x}{-3}=\frac{27}{x-2}\)
\(\Leftrightarrow\left(2-x\right)\left(x-2\right)=-3\cdot27\)
\(\Leftrightarrow-\left(x-2\right)\left(x-2\right)=-81\)
\(\Leftrightarrow-\left(x-2\right)^2=-81\)
\(\Leftrightarrow\left(x-2\right)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=9\\x-2=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-7\end{cases}}\)