a) - Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

    + Ta có: \(A=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

      \(\Leftrightarrow A=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

    - Đặt \(a=x^2+7x+10\)

    + Ta lại có: \(A=a.\left(a+2\right)-24\)

               \(\Leftrightarrow A=a^2+2a-24\)

               \(\Leftrightarrow A=\left(a^2-4a\right)+\left(6a-24\right)\)

               \(\Leftrightarrow A=a.\left(a-4\right)+6.\left(a-4\right)\)

               \(\Leftrightarrow A=\left(a-4\right).\left(a+6\right)\)

    - Thay \(a=x^2+7x+10\)vào phương trình \(A\), ta có:

                     \(A=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

              \(\Leftrightarrow A=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

              \(\Leftrightarrow A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

^_^ Chúc bạn hok tốt ^_^ !!#@##

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

\(\Leftrightarrow x^3+2x^2+x+x^2+2x+1-4=x^3+3x^2\)

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

\(\Leftrightarrow3x^2+3x-3=3x^2\)

\(\Leftrightarrow3x-3=0\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

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

\(=>x^3+2x^2+x+x^2+2x+1-4=x^3+3x^2\)

\(=>x^3+3x^2+3x-3=x^3+3x^2\)

\(=>3x-3=0\)

\(=>3.\left(x-1\right)=0\)

\(=>x-1=0\)

\(=>x=1\)

a) Câu hỏi của a - Toán lớp 8 - Học toán với OnlineMath

b) Câu hỏi của c - Toán lớp 8 - Học toán với OnlineMath

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

        \(\Leftrightarrow A=x^2.\left(x+1\right)-4.\left(x-1\right).\left(x+1\right)\)

        \(\Leftrightarrow A=\left(x+1\right).\left[x^2-4.\left(x-1\right)\right]\)

        \(\Leftrightarrow A=\left(x+1\right).\left(x^2-4x+4\right)\)

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

\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\left(3x+2\right)\left[3x^2+9x+13-\left(9x^2-6x+4\right)\right]=0\)

\(\left(3x+2\right)\left[3x^2+9x+13-9x^2+6x-4\right]=0\)

TH1: \(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow\frac{-2}{3}\)

TH2: \(3x^2+9x+13+9x^2+6x-4=0\)

\(\Leftrightarrow-6x^2+15x+9=0\)

\(\Leftrightarrow\left(-6x-3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-6x-3=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-6x=3\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=3\end{cases}}}\)

\(A=\left(x+y+z+\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4z}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\ge2\sqrt{x.\frac{1}{4x}}+2\sqrt{y.\frac{1}{4y}}+2\sqrt{z.\frac{1}{4z}}+\frac{3}{4}\left(\frac{9}{x+y+z}\right)\)

\(\ge1+1+1+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)

Dấu "=" xảy ra <=> x = y = z = 1/2

Vậy min A = 15/2 tại x = y = z = 1/2

Lời giải của em ạ :D

\(A=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\ge x+y+z+\frac{9}{x+y+z}\)

Đặt \(t=x+y+z\le\frac{3}{2}\)

Khi đó \(A=t+\frac{9}{t}=\left(t+\frac{9}{4t}\right)+\frac{27}{4t}\ge3+\frac{27}{4\cdot\frac{3}{2}}=\frac{15}{2}\)

Đẳng thức xảy ra tại x=y=z=¹/₂

đkxđ \(x\ne1\)

\(\Leftrightarrow\left(m-1\right)\left(x-1\right)=2m-2\)

\(\Leftrightarrow mx-m-x+1=2m-2\)

\(\Leftrightarrow mx-x=3m-3\)

\(\Leftrightarrow x\left(m-1\right)=3\left(m-1\right)\)(*)

Biện luận

+ Nếu m = 1 pt (*) 0x = 0 (vsn)

+ Nếu m khác 1 pt (*) -2x = -6 (cn)

Kết luận m khác 1 thì pt có nghiệm

               m=1 thì pt vsn

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\left(x\ne\pm3\right)\)

\(\Leftrightarrow\frac{x+3}{x-3}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{x+3}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{x^2+6x+9-17-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{12x-17}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow12x-17=0\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\frac{17}{12}\left(tmđk\right)\)