\(9x^2-1+\left(3x-1\right).\left(x+2\right)=0\)

\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)

\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)

\(\Leftrightarrow12x^2+5x=3\)

\(\Leftrightarrow12x^2+5x-3=0\)

\(\Leftrightarrow12x^2-4x+9x-3=0\)

\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)

\(\Leftrightarrow\left(4x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)

tách nhỏ câu hỏi ra

1. -3(-x+3)

= 3x - 6

2. -5x³ (-3x + 5)

= 15x⁴ - 25x³

3. -2x (-2x - 6)

= 4x² + 12x

ĐK : \(\begin{cases}x\ge\frac{-1}{3}\\y\le5\end{cases}\)

\(\sqrt{5x^2+3y+1}+1-4x=0\)

\(\Leftrightarrow\begin{cases}x\ge\frac{1}{4}\\5x^2+3y+1=16x^2-8x+1\left(1\right)\end{cases}\)

(1) \(\Leftrightarrow11x^2-8x-3y=0\left(2\right)\)

Đặt \(\begin{cases}\sqrt{3x+1}=a\left(a\ge0\right)\\\sqrt{5-y}=b\left(b\ge0\right)\end{cases}\) \(\Rightarrow\begin{cases}3x+2=a^2+1\\6-y=b^2+1\end{cases}\)

\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\\ \Leftrightarrow a^3-b^3+a-b=0\\ \Leftrightarrow\left(a-b\right)\left(a^2-ab+b^2+1\right)=0\\ \Leftrightarrow a-b=0\left(a^2-ab+b^2+1>0\right)\\\Leftrightarrow a=b\\ \)

\(\Rightarrow\sqrt{3x+1}=\sqrt{5-y}\\ \Leftrightarrow3x+1=5-y\\ \Leftrightarrow y=4-3x\left(3\right)\)

Từ (2) và (3)

 \(\Rightarrow11x^2-8x-3\left(4-3x\right)=0\\ \Leftrightarrow11x^2+x-12=0\\ \Leftrightarrow x=1\left(TM\right);x=\frac{-12}{11}\left(loại\right)\\ \Rightarrow y=1\left(TM\right)\)

Vậy S = \(\left\{\left(1;1\right)\right\}\)

no biết

Mình mới lớp 6 Sorry

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

\(\Leftrightarrow\dfrac{x^2-9}{3}+\dfrac{6}{3}=\dfrac{3x\left(1-x\right)}{3}\)

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

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

\(\Leftrightarrow4x^2-3x-3=0\)

\(\Delta=9-4\cdot4\cdot\left(-3\right)=9+48=57\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{57}}{8}\\x_2=\dfrac{3+\sqrt{57}}{8}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3-\sqrt{57}}{8};\dfrac{3+\sqrt{57}}{8}\right\}\)

Tớ học ngu nên chỉ biết cách nhân ra rồi rút gọn chứ không biết cách nào ngắn hơn :)) Hơi dài dòng nên phân tích từng vế 1 nhé :D

2/ \(\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)=4\left(2x^2+5x-204\right)\left(x^2-5x-206\right)\)

*****\(VT=\left(2x^2+5x-204\right)^2+4\left(x^2-5x-206\right)^2\)

\(=4x^4+25x^2+41616+20x^3-816x^2-2040x+4\left(x^4-387x^2+42436-10x^3+2060x\right)\)

\(=4x^2+25x^2+41616+20x^3-816x^2-2040x+4x^2-1548x^2+169744-40x^3+8240x\)

\(=8x^4-1523x^2+6200x+211360\)

*****\(VP=\left(8x^2+20x-816\right)\left(x^2-5x-206\right)\)

\(=8x^4-40x^3-1648x^2-100x^2-4120x-816x^2+4080x+168096\)

\(=8x^4-1748x^2-40x+168096\)

\(\Rightarrow8x^4-1523x^2+6200x+211360=8x^4-1748x^2-40x+168096\)

\(\Leftrightarrow-1523x^2+6200x+211360+1748x^2-40x+168096=0\)

\(\Leftrightarrow255x^2+43264+6240x=0\)

\(\Leftrightarrow\left(15x+208\right)^2=0\)

\(\Leftrightarrow15x+208=0\)

\(\Leftrightarrow x=-\frac{208}{15}\)

+ Ta có: \(x^4-5x^3+6x^2+5x+1=0\)

        \(\Rightarrow x^2-5x+6+\frac{5}{x}+\frac{1}{x^2}=0\)( chia cả hai vế cho \(x^2\))

       \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-\left(5x-\frac{5}{x}\right)+6=0\)

      \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-5.\left(x-\frac{1}{x}\right)+6=0\)( *** )

- Đặt  \(x-\frac{1}{x}=a\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=a^2+2\)

- Thay  \(a=x-\frac{1}{x};\)\(a^2+2=x^2+\frac{1}{x^2}\)vào ( *** )

- Ta có: \(a^2+2-5a+6=0\)

     \(\Leftrightarrow a^2-5a+8=0\)

     \(\Leftrightarrow4a^2-20a+32=0\)

     \(\Leftrightarrow\left(4a^2-20a+25\right)+7=0\)

     \(\Leftrightarrow\left(2a-5\right)^2+7=0\)

- Ta lại có: \(\hept{\begin{cases}\left(2a-5\right)^2\ge0\forall a\\7>0\end{cases}}\Rightarrow \left(2a-5\right)^2+7\ge7>0\)mà \(\left(2a-5\right)^2+7=0\)

\(\Rightarrow\left(2a-5\right)^2+7\)( vô nghiệm ) \(\Rightarrow\)\(x^4-5x^3+6x^2+5x+1=0\)( vô nghiệm )

Vậy \(S=\left\{\varnothing\right\}\)

+ Ta có: \(\left(2x^2+5x-204\right)^2+4.\left(x^2-5x-206\right)=4.\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)( ** )

- Đặt \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)

- Thay \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)\(\Rightarrow\)\(a.b=\left(2x^2+5x-204\right).\left(x^2-5x-206\right)\)

vào ( ** )

- Ta có: \(a^2+4b^2=4ab\)

      \(\Leftrightarrow a^2-4ab+4b^2=0\)

      \(\Leftrightarrow\left(a-2b\right)^2=0\)

      \(\Leftrightarrow a-2b=0\)

      \(\Leftrightarrow a=2b\)( * )

- Thay  \(a=2x^2+5x-204;\)\(b=x^2-5x-206\)vào ( * )

- Ta lại có: \(2x^2+5x-204=2.\left(x^2-5x-206\right)\)

       \(\Leftrightarrow2x^2+5x-204=2x^2-10x-412\)

      \(\Leftrightarrow\left(2x^2-2x^2\right)+\left(5x+10x\right)=-\left(412-204\right)\)

      \(\Leftrightarrow15x=-208\)

      \(\Leftrightarrow x=-\frac{208}{15} \left(TM\right)\)

Vậy \(S=\left\{-\frac{208}{15}\right\}\)