A) (15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52

..............bn phân rồi gộp lại để ra kq như dòng dưới nha....

=>19x + 56 = 52

=> 19x = -4

=> x = ‐ 4 / 1 9

NHỚ TK MK ĐÓ

bằng 6849/5000 nha

pt <=> ( 2x + 3 )( x - 5 ) - 2x( 2x + 3 ) = 0

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

<=> x = -3/2 hoặc x = -5

Vậy ... 

\(\left(2x+3\right)\left(x-5\right)=4x^2+6x\Leftrightarrow\left(2x+3\right)\left(x-5\right)=2x\left(2x+3\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{2};x=-5\)

Vậy tập nghiệm của pt là S = { -5 ; -3/2 } 

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

\(=8\left(2x+3\right)=16\)

\(\Rightarrow2x+3=2\)

\(\Rightarrow x=-\frac{1}{2}\)

1)243/-(3)^2x=3

=>243:3=(-3)^2x=81=9^2=3^2^2=(-3)^2^2=(-3)^2.2

=>2x=2.2=>x=2

2/ 25^x/(-5)^4=125^2

=5^2.x/5^2.5^2=5^3^2

=x/5^2=5^6

=>x=5^6.5^2=5^8

nếu đúng thì

Tách tách tách :v

$(15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52$

$=>(60x+15-8x^2-2x)-(26x-39-8x^2+12x)-(x^2+3x+2)+x^2=52$

$=>60x+15-8x^2-2x-26x+39+8x^2-12x-x^2-3x-2+x^2=52$

$=>(8x^2-8x^2+x^2-x^2)+(60x-2x-26x-12x-3x)+(15+39-2)=52$

$=>17x+52=52$

$=>x=0$

\(3+\sqrt{2x-3}=x\) (ĐKXĐ: x \(\ge\)1,5)

\(\Leftrightarrow\sqrt{2x-3}=x-3\)

\(\Leftrightarrow2x-3=x^2-6x+9\)

\(\Leftrightarrow-x^2+8x-12=0\)

\(\Leftrightarrow-\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow x^2-6x-2x+12=0\)

\(\Leftrightarrow x.\left(x-6\right)-2.\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\x=2\end{cases}\left(\text{TMĐK}\right)}\)

Vậy ...

a:

ĐKXĐ: x<>2

|2x-3|=1

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

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)

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

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

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

\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)

c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)

\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)