a: =>2x+5=4

=>2x=-1

hay x=-1/2

b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)

=>(3x-4)(3x-5)(3x-3)=0

hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)

c: \(\Leftrightarrow3^{x+1}=3^{2x}\)

=>2x=x+1

=>x=1

d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)

=>2x+3=2x-10

=>0x=-13(vô lý)

mk tìm min bn h nha, chịu k?

thôi bn k h mk cũng làm;

= x² - 2,3x/16 + (3/16)² - 9/256 + 503/256

=(x+3/16)² + 494/256

min = 494/256

Cách kia hơi lằng nhằng.

Làm lại:

\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}=\left(-\frac{3}{4}\right)^{-4}\)

\(\Rightarrow3x-1=-4\)

\(\Rightarrow3x=-3\)

\(\Rightarrow x=-1\)

Ta có : \(\left(-\frac{3}{4}\right)^{-4}=\frac{256}{81}\)

do đó 3x - 1 = -4

=> 3x = -4 + 1 = -3

=> x = -3 : 3 = -1

\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\Rightarrow\left(-\frac{3}{4}\right)^{3x}:\left(-\frac{3}{4}\right)=\frac{256}{81}\Rightarrow\left(-\frac{3}{4}\right)^{3x}=-\frac{64}{27}\Rightarrow\left(-\frac{3^3}{4^4}\right)^x=-\frac{64}{27}\Rightarrow\left(-\frac{27}{64}\right)^x=-\frac{64}{27}\Rightarrow x=-1\)

a)=18-5-7

=13-7

=5

b)=(48-2)+(-21)

46+(-21)

=25

c)=-200.(-2)+20

=400+20=420