Chỗ câu b ý, 3³ = 27 mà ta :))

🍀🧡_Trang_🧡🍀 mình lộn ý

Tìm GTNN của biểu thức :

\(x^2+2x+4\)

Đặt A = \(x^2+2x+4\)

\(\Leftrightarrow A=\left(x^2+2.x.1+1\right)+3\)

\(\Leftrightarrow A=\left(x+1\right)^2+3\)

Ta luôn có : \(\left(x+1\right)^2\ge0\forall x\)

Suy ra : \(\left(x+1\right)^2+3\ge3\forall x\)

Hay A\(\ge3\) với mọi x

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Nên : \(A_{min}=3khix=-1\)

m: \(=\left(\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{2\left(x+1\right)}\right)\cdot\dfrac{2x}{x+1}-\dfrac{3}{x-1}\)

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

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

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

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

\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)

Tìm x

a) Ta có: \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)-15=0\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow40x=40\)

hay x=1

Vậy: x=1

b) Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)-49=0\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

hay x=3

Vậy: x=3

d) Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

hay \(x=\frac{5}{12}\)

Vậy: \(x=\frac{5}{12}\)

e) Ta có: \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x-9=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow6x=16\)

hay \(x=\frac{8}{3}\)

Vậy: \(x=\frac{8}{3}\)

f) Ta có: \(\left(x-5\right)^2-\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25-\left(x-x^2-4+4x\right)=0\)

\(\Leftrightarrow x^2-10x+25-x+x^2+4-4x=0\)

\(\Leftrightarrow2x^2-15x+29=0\)

\(\Leftrightarrow2\left(x^2-\frac{15}{2}x+\frac{29}{2}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{15}{4}+\frac{225}{16}+\frac{7}{16}=0\)

\(\Leftrightarrow\left(x-\frac{15}{4}\right)^2+\frac{7}{16}=0\)(vô lý)

Vậy: x∈∅

\(a,\left(x+1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)\(\Leftrightarrow x^3+3x^2+3x+1+8-x^3+3x^2+6x-17=0\)\(\Leftrightarrow6x^2+9x-8=0\)

\(\Leftrightarrow x^2+\dfrac{3}{2}x-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{16}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2=\dfrac{91}{48}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\sqrt{\dfrac{91}{48}}\\x+\dfrac{3}{4}=-\sqrt{\dfrac{91}{48}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\\x=-\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{273}}{12}\\x=-\dfrac{9+\sqrt{273}}{12}\end{matrix}\right.\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

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

A = \(x^2-6x+9-4x^2+1=-3x^2-6x+10\)

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

B = \(4x^2-12x+9-2x^2-x+2x+1\)

B = \(2x^2-11x+10\)

C = \(4x\left(x-3\right)^2-\left(4-2x\right)^2\)

C = \(4x\left(x^2-6x+9\right)-16+16x-4x^2\)

C = \(4x^3-24x^2+36x-16+16x-4x^2\)

C = \(4x^3-28x^2+52x-16\)

D = \(3x\left(x-1\right)\left(x-2\right)-x\left(2x-1\right)^2\)

D = \(\left(3x^2-3x\right)\left(x-2\right)-x\left(2x-1\right)^2\)

D = \(3x^3-6x^2-3x^2+6x-x\left(4x^2-4x+1\right)\)

D = \(3x^3-9x^2+6x-4x^3+4x^2-x\)

D = \(-x^3-5x^2+5x\)

Đáp án câu C cho sẵn là:C=4x-16 bn ạ

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

x^2+2x-3/3+2x/4=x^2/3