x=-370

`(2x+5)(2x-7)-(2x-3)^2=36`

`<=>4x^2-14x+10x-35-(4x^2-12x+9)=36`

`<=>4x^2-4x-35-4x^2+12x-9=36`

`<=>8x-44=36`

`<=>8x=80`

`<=>x=10`

Vậy `S={10}`

Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(2x-3\right)^2=36\)

\(\Leftrightarrow4x^2-14x+10x-35-\left(4x^2-12x+9\right)=36\)

\(\Leftrightarrow4x^2-4x-35-4x^2+12x-9=36\)

\(\Leftrightarrow8x-44=36\)

\(\Leftrightarrow8x=80\)

hay x=10

Vậy: S={10}

Giúp tớ với tớ cần gấp

a: =>4x-15-2x+3=36

=>2x-12=36

=>2x=48

hay x=24

b: =>2x-x=5+4

hay x=9

a. (2x-3)² = 36

(2x-3)² = 6²

=> TH1: 2x - 3 = 6

2x = 9

x = 9/2

TH2: 2x - 3 = -6

2x = -6 + 3

2x = -3

x = -3/2

Vậy x \(\in\){ -3/2 ; 9/2)

Câu b tương tự

a.(2x-3)^2=36

\(\Rightarrow\left(2x-3\right)^2=6^2\)

\(\Rightarrow2x-3=6\)

\(\Rightarrow2x=9\)\(\Rightarrow x=9:2=\frac{9}{2}\)

a)\(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=6^2\)

\(\Rightarrow\hept{\begin{cases}2x-3=6\\2x-3=-6\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{2}\\x=-\frac{3}{2}\end{cases}}}\)

b)

ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

Để A đạt giá trị nguyên thì \(-2x^2+x+36⋮2x+3\)

\(\Leftrightarrow-2x^2-3x+4x+6+30⋮2x+3\)

\(\Leftrightarrow-x\left(2x+3\right)+2\left(2x+3\right)+30⋮2x+3\)

\(\Leftrightarrow\left(2x+3\right)\left(-x+2\right)+30⋮2x+3\)

mà \(\left(2x+3\right)\left(-x+2\right)⋮2x+3\)

nên \(30⋮2x+3\)

\(\Leftrightarrow2x+3\inƯ\left(30\right)\)

\(\Leftrightarrow2x+3\in\left\{1;-1;2;-2;3;-3;5;-5;6;-6;10;-10;15;-15;30;-30\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;1;-5;0;-6;2;-8;3;-9;7;-13;12;-18;27;-33\right\}\)

hay \(x\in\left\{-1;-2;\dfrac{1}{2};\dfrac{-5}{2};0;-3;1;-4;\dfrac{3}{2};\dfrac{-9}{2};\dfrac{7}{2};\dfrac{-13}{2};6;-9;\dfrac{27}{2};\dfrac{-33}{2}\right\}\)(thỏa ĐK)

Vậy: \(x\in\left\{-1;-2;\dfrac{1}{2};\dfrac{-5}{2};0;-3;1;-4;\dfrac{3}{2};\dfrac{-9}{2};\dfrac{7}{2};\dfrac{-13}{2};6;-9;\dfrac{27}{2};\dfrac{-33}{2}\right\}\)

\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)

a: Ta có: \(\left(x+2\right)^2+\left(2x-1\right)^2-\left(x-3\right)^2=36\)

\(\Leftrightarrow x^2+4x+4+4x^2-4x+1-x^2+6x-9=36\)

\(\Leftrightarrow4x^2+6x-4-36=0\)

\(\Leftrightarrow4x^2+6x-40=0\)

\(\text{Δ}=6^2-4\cdot4\cdot\left(-40\right)=676\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-6-26}{8}=-4\\x_2=\dfrac{-6+26}{8}=\dfrac{5}{2}\end{matrix}\right.\)

\(=\left(x^3+4x^2-6x^2-24x+9x+36\right):\left(x+4\right)\\ =\left[x^2\left(x+4\right)-6x\left(x+4\right)+9\left(x+4\right)\right]:\left(x+4\right)\\ =x^2-6x+9\)

`2x -2/3 +1/2x =-1`

`=> 2x+1/2x =-1+2/3`

`=> (2+1/2) x =-3/3 + 2/3`

`= 5/2 x = -1/3`

`=> x=-1/3 : 5/2`

`=>x= -2/15`

`31/36 - (1/3-x)^2 =5/6`

`=>  (1/3-x)^2 = 31/36 - 5/6`

`=>  (1/3-x)^2 =1/36`

`=>  (1/3-x)^2 = (+-1/6)^2`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{1}{6}\\\dfrac{1}{3}-x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)