b: Ta có: \(B=-2x^2+4x+1\)

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

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

\(=-2\left(x-1\right)^2+3\le3\forall x\)

Dấu '=' xảy ra khi x=1

Tôi học lớp 3 ok

jmkhgfda????

tui học lớp 4 

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)

(2\(x\) + 3)² + (3\(x\) - 2)⁴  =0

Vì:

(2\(x\) + 3)² ≥ 0

(3\(x\) - 2)⁴ ≥ 0

Nên :

(2\(x\) + 3)² + (3\(x\) - 2)⁴ = 0

⇔ \(\left\{{}\begin{matrix}2x+3=0\\3x-2=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) \(\varnothing\)

1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
suy ra 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
suy ra 1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
suy ra 1 - 1/2x = 2/8
suy ra 1/2x = 1 - 2/8
suy ra 1/2x = 6/8 = 3/4
suy ra 1.4 = 2.x.3
suy ra 4 = 6x
suy ra x thuộc rỗng
 Vậy x thuộc rỗng
k cho mình nha. Chúc bạn học tốt!

Thanks you so much.

Thùy Anh tính sai rùi, dòng 2 phải nhân cả tử và mẫu cùng zới 2 mò, ko nhận ra ak mng ưi

Ta có:

\(\left(2x-3\right)\left(2x+3\right)=x\left(x-3\right)\)

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

\(\Rightarrow4x^2-9=x^2-3x\)

\(\Rightarrow4x^2-x^2+3x=9\)

\(\Rightarrow3x^2+3x=9\)

\(\Rightarrow3\left(x^2+x\right)=9\)

\(\Rightarrow x^2+x=3\)

\(\Rightarrow x^2+2\times\frac{1}{2}x+\frac{1}{4}=3+\frac{1}{4}=\frac{13}{4}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{13}{4}\)

Đến đây bạn tự làm nốt nhá!

sao lại cộng vs 1/4 zậy

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)

\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)

\(c,\\ \left(2x+1\right)^3=9.81\\ \left(2x+1\right)^3=3^2.3^4\\ \left(2x+1\right)^3=3^6\\ \left(2x+1\right)^3=\left(3^2\right)^3=9^3\\ Vậy:2x+1=9\\ 2x=9-1\\ 2x=8\\ x=\dfrac{8}{2}\\ x=4\)

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)