\(a,\left(2x-16\right)^7=128\\ \Rightarrow\left(2x-16\right)^7=2^7\\ \Rightarrow2x-16=2\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ b,x^3.x^2=2^8:2^3\\ \Rightarrow x^5=2^5\\ \Rightarrow x=5\\ c,3^{x-3}-3^2=2.3^2\\ \Rightarrow3^{x-3}-9=18\\ \Rightarrow3^{x-3}=27\\ \Rightarrow3^{x-3}=3^3\\ \Rightarrow x-3=3\\ \Rightarrow x=6\)

D

2.x-15=x+11-(-16)

2.x-15=x+27

\(2x-15=x+11-\left(-16\right)\)

\(\Rightarrow2x-x=11+16+15\)

\(\Rightarrow x=27+15\)

\(\Rightarrow x=42\)

Vậy..................

2x  - 15 = x +11- ( -16 )

2x  - 15 = x + 27

2x - x  = 27 - 15

x         = 12

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y²-2y+1=0⇒ (y-1)²=0⇒y-1=0⇒y=1

Do đó Q=x²+y²=(-1)²+1²=2

\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

d) Ta có: \(x^2-6x-16=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

3,2 + 4,65 : 1,5 

=3,2+3,1

=6,3

347 + 418 + 123 + 12

=(347+123)+(418+12)

=470+430

=900

34 x 35 + 35 x 38 + 65 x 40 + 65 x 32

=35*(34+38)+65*(32+40)

=35*72+65*72

=72*(35+65)

=72*100

=7200

19 x 16 x 4 + 76 x 34

=76*16+76*34

=76*(16+34)

=76*50

=3800

136 x 68 + 16 x2 x 136 

=136*68+32*136

=136*(68+32)

=136*100

=13600

125 x 25 + 125 x 74 + 125

=125*(25+74+1)

=125*100

=12500

Tính bằng cách thuận tiện hả

3,2 + 4,65 : 1,5 = 3,2 + 3,1 = 6,3

347 + 418 + 123 + 12 = (347 + 123) + (418 + 12) = 470 + 430 = 900

34 x 35 + 35 x 38 + 65 x 40 + 65 x 32 = 35 x ( 34 + 38) + 65 x (40 + 32) = 35 x 72 + 65 x 72 = 72 x (35 + 65) = 72 x 100 = 7200

19 x 16 x 4 + 76 x 34

136 x 68 + 16 x2 x 136 

125 x 25 + 125 x 74 + 125

\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)

\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

x=5 và 0