Đáp án C

Đáp án : C

\(a.\left(x+5\right)^3=-64\Leftrightarrow x+5=-4\Leftrightarrow x=-9\\ b.\left(2x-3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\) 

( Nhớ kết luận! )

a.

\(\left(x+5\right)^3=\left(-4\right)^3\)

\(\Leftrightarrow x+5=-4\)

\(\Leftrightarrow x=-9\)

b.

\(\left(2x-3\right)^2=3^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

\(\left(x+5\right)^3=\left(-4\right)^3\)

=-9

Vậy 

(9 không được)

\(\left(2x-3\right)^3=2^3\)

P/s Nhớ tick cho mình nha. Thanks bạn

1.

a) \(2^x=128\)

\(2^x=2^7\)

\(=>x=7\)

b) \(8^{x-1}=64\)

\(8^{x-1}=8^2\)

\(=>x-1=2\)

\(x=2+1\)

\(=>x=3\)

c) \(3+3^x=30\)

\(3^x=30-3\)

\(3^x=27=3^3\)

\(=>x=3\)

d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?

e) \(3^2.x=3^5\)

\(x=3^5:3^2\)

\(=>x=3^3=27\)

f) \(\left(2x-1\right)^3=343\)

\(\left(2x-1\right)^3=7^3\)

\(=>2x-1=7\)

\(2x=7+1\)

\(2x=8\)

\(x=8:2\)

\(=>x=4\)

\(#Wendy.Dang\) 

a,\(2^x\)=128           b,\(8^{x-1}\)=64              c,3+\(3^x\)=30           d,x+2=64

\(2^7\)=128               \(8^{x-1}\)=\(8^2\)                 \(3^x\)=30-3                  x=64-2

=>x=7              =>x-1=2                  \(3^x\)=27                      x=62

                         x=2+1=3                \(3^x\)=\(3^3\)

                                                     =>x=3

e,\(3^2\).x=\(3^5\)                             f,(2x-\(1^3\))=343

x=\(3^5\):\(3^2\)                                 2x=1+343

x=27                                     2x=344

                                               x=344:2

                                               x=172

a: M(x)=x^2-16x+64=(x-8)^2

Đặt M(x)=0

=>x-8=0

=>x=8

\(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow x-3=\left(x-3\right)^2\)

\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

___________

\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

a) \(3x^3-6x^2=0\)

\(3x^2\left(x-2\right)=0\)

\(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) \(x\left(x-4\right)-12x+48=0\)

\(x^2-4x-12x+48=0\)

\(x^2-16x+48=0\)

\(\left(x-12\right)\left(x-4\right)=0\)

\(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)

\(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)

c) Viết thiếu nha :v

d) \(2x\left(x-5\right)-x\left(2x+3\right)=16\)

\(2x^2-10x-x^2-2x^2-3x=16\)

\(-13x=16\)

\(x=-\frac{16}{13}\)

e) \(\left(4x^2-1\right)-\left(x-1\right)^2=-3\)

\(4x^2-1-x^2+2x-1=-3\)

\(3x^2-2+2x=-3\)

\(3x^2-2+2x+3=0\)

\(3x^2+1+2x=0\)

Vì \(3x^2+1+2x>0\)nên: 

\(x\in\varnothing\)

A) 3x³ - 6x² = 0

=> 3x²(x - 2) = 0

=> \(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b) x(x - 4) - 12x + 48 = 0

=> x(x - 4) - 12(x - 4) = 0

=> (x - 12)(x - 4) = 0

=> \(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)

c) x(x - 4) - (x² - 8) = x² - 4x - x² + 8 = 4x + 8 

\(a,3x^3-6x^2=0\Rightarrow3x^2\left(x-2\right)=0.\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

\(b,x\left(x-4\right)-12x+48=0\)

\(\Rightarrow x\left(x-4\right)-12\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-12=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=12\end{cases}}}\)

\(c,x\left(x-4\right)-\left(x^2-8\right)=0\)

\(\Rightarrow c,x^2-4x-x^2+8=0\)

\(\Rightarrow-4x+8=0\)

\(\Rightarrow-4\left(x-2\right)=0\)

\(\Rightarrow x=2\)

\(d,2x\left(x-5\right)-x\left(2x+3\right)=16\)

\(\Rightarrow2x^2-10x-2x^2-3x=16\)

\(\Rightarrow-13x=16\Leftrightarrow x=-\frac{16}{13}\)

\(e,\left(4x^2-1\right)-\left(x-1\right)^2=-3\)

\(\Rightarrow4x^2-1-x^2+2x-1=-3\)

\(\Rightarrow3x^2+2x+1=0\)

\(\Rightarrow x^2+\frac{2}{3}x+\frac{1}{3}=0\)

\(\Rightarrow x^2+2.x.\frac{1}{3}+\frac{1}{9}+\frac{2}{9}=0\)

\(\Rightarrow\left(x+\frac{1}{3}\right)^2+\frac{2}{9}=0\)( vô lý ) 

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