\(16^x+7.4^x+5=3.2^x+2\)

<=> \(8.2^x+7.2.2^x+5=3.2^x+2\)

<=> \(8.2^x+7.2.2^x+5-3.2^x-2=0\)

<=> \(2^x\left(8+7.2-3\right)-3=0\)

<=> \(2^x.19=3\) 

<=> \(2^x=\frac{3}{19}\)

\(16^x=2^x.8^x\) mà

a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)

c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)

a)\(\Leftrightarrow\)\(x^2-4x-21=0\)

\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)

\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)

\(\Leftrightarrow\)\((x-7)(x+3)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)

b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)

\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)

\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)

c)\((x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15\)

\(\Leftrightarrow\)\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9-15=0\)

\(\Leftrightarrow\)\(45x-6=0\)

\(\Leftrightarrow\)\(x=\dfrac{2}{15}\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a, 0,5x.(2x - 9) = 1,5x.(x - 5)

<=> x² - 4,5x = 1,5x² - 7,5x

<=> 0,5x² + 3x = 0

<=> 0,5x.( x + 6 ) = 0

<=> x = 0 hoặc x + 6 = 0

<=> x = 0 hoặc x = -6

Vậy....

#Đức Lộc#

Làm thử nha :v

a) 0,5x.(2x - 9) = 1,5x.(x - 5)

<=> 0,5x.(2x - 9) - 1,5x.(x - 5) = 1,5x.(x - 5) - 1,5x.(x - 5)

<=> 0,5x.(2x - 9) - 1,5x.(x - 5) = 0

<=> -x(0,5x - 3) = 0

=> x = 0 hoặc 6

b) 5(x - 1) - (2x - 5) = 16 - x

<=> 3x = 16 - x

<=> 3x + x = 16

<=> 4x = 16

<=> x = 16 : 4

=> x = 4

a) \(0,5x\left(2x-9\right)=1,5x\left(x-5\right)\Leftrightarrow x^2-4,5x=1,5x^2-7,5x\)

\(\Leftrightarrow x^2-1,5x^2=4,5x-7,5x\Leftrightarrow-0.5x=-3x\Leftrightarrow x=6\)

Vậy phương trình có tập nghiệm S = { 6 }

 b) \(5\left(x-1\right)-\left(2x-5\right)=16-x\Leftrightarrow5x-5-2x+5=16-x\)

\(\Leftrightarrow5x-2x+x=5-5+16\Leftrightarrow4x=16\Leftrightarrow x=4\)

Vậy phương trình có tập nghiệm S = { 4 }

d)\(-ĐKXĐ:\hept{\begin{cases}x+1\ne0\\x-2\ne0\\\left(x+1\right)\left(x-2\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Ta có:  \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\Leftrightarrow\frac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\left(x+2\right)}=\) \(\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow2x-x+3x=4+1-11\Leftrightarrow4x=-6\Leftrightarrow x=-\frac{3}{2}\)

Vậy phương trình có tập nghiệm S = { -3/2 }

a: ĐKXĐ: x>=-2

\(PT\Leftrightarrow3\cdot3\sqrt{x+2}=\dfrac{1}{2}\cdot2\sqrt{x+2}+16\)

=>\(9\sqrt{x+2}-\sqrt{x+2}=16\)

=>\(8\sqrt{x+2}=16\)

=>\(\sqrt{x+2}=2\)

=>x+2=4

=>x=2

b: ĐKXĐ: \(x\in R\)

\(5+\sqrt{x^2-4x+4}=9\)

=>\(\left|x-2\right|=4\)

=>x-2=4 hoặc x-2=-4

=>x=6 hoặc x=-2

a) 16 - 3x = 4

<=> 3x = 12

<=> x = 4

Vậy x = 4 là nghiệm phương trình 

b) (x² - 4x + 5)² - (x - 1)(x - 3) = 4

<=> (x² - 4x + 5)² - 4 - (x - 1)(x - 3) = 0

<=> (x² - 4x + 5 - 2)(x² - 4x + 5 + 2) - (x - 1)(x - 3) = 0

<=> (x² - 4x + 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 6) = 0

<=> (x  - 1)(x - 3) = 0 (Vì x² - 4x + 6 > 0 \(\forall x\))

<=> \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy x \(\in\left\{1;3\right\}\)là nghiệm phương trình 

a)16-3x=4

3x=16-4

3x=12

x=4

Vậy x=4

b)(x²-4x+5)²-(x-1).(x-3)=4

[(x-2)²+1]²-[(x-2)+1].[(x-2)-1]=4

=>(x-2)²+2.(x-2).1+1-(x-2)²-1²=4

2(x-2)=4

=>x-2=2

=>x=4

Vậy ....................

Chú bn học tốt

a) 16 - 3x = 4

<=> 3x = 12

<=> x = 4

Vậy x = 4 là nghiệm phương trình 

b) (x² - 4x + 5)² - (x - 1)(x - 3) = 4

<=> (x² - 4x + 5)² - 4 - (x - 1)(x - 3) = 0

<=> (x² - 4x + 5 - 2)(x² - 4x + 5 + 2) - (x - 1)(x - 3) = 0

<=> (x² - 4x + 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 7) - (x - 1)(x - 3) = 0

<=> (x - 1)(x - 3)(x² - 4x + 6) = 0

<=> (x  - 1)(x - 3) = 0 (Vì x² - 4x + 6 > 0 ∀x)

<=> [

⇔[

Vậy x ∈{1;3}là nghiệm phương trình 

a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)

b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)

\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0 

<=>  (x - 3)(4x - 1 - 5x - 2) = 0

<=>  (x - 3)(-x - 3) = 0

<=>  x  = 3 hoặc x = -3

b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0 

<=>  (x + 3)(x - 5 + 3x - 4) = 0

<=>  (x + 3)(4x - 9) = 0

<=>  x = -3 hoặc x = 9/4

c) ( x + 6 ) ( 3x - 1 )+ x² - 36 = 0 

<=>  3x^2 + 17x - 6 + x^2 - 36 = 0

<=>  4x^2 + 17x - 42 = 0

<=>  4x^2 + 24x - 7x - 42 = 0

<=>  4x(x + 6) - 7(x + 6) = 0

<=>  (4x - 7)(x + 6) = 0

<=>  x = -6 hoặc x = 7/4

d) ( x + 4 ) ( 5x + 9 ) - x² + 16 = 0 

<=>  5x^2 + 29x + 36 - x^2 + 16 = 0

<=>  4x^2 + 29x + 52 = 0

<=>  4x^2 + 16x + 13x + 42 = 0

<=>  4x(x + 4) + 13(x + 4) = 0

<=>  (4x + 13)(x + 4) = 0

<=>  x = -13/4 và x = -4