\(\Leftrightarrow\left(x^2-12x-6\right)\left(x^2-4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2=42\\\left(x-2\right)^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6\in\left\{\sqrt{42};-\sqrt{42}\right\}\\x-2\in\left\{\sqrt{2};-\sqrt{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{42}+6;-\sqrt{42}+6;\sqrt{2}+2;2-\sqrt{2}\right\}\)

Đây đích thực có phải là lớp 1 ko bn?

Lên lớp trên mà gửi đi gửi ở lớp 1 làm gì vậy bạn .

Kiểm tra lại đề câu a, \(...+24\) thì pt vô nghiệm, phải là \(...-24\) mới có lý

b/ \(x^2-\left(y+1\right)x+y^2-y-2=0\) (1)

\(\Delta=\left(y+1\right)^2-4\left(y^2-y-2\right)\ge0\)

\(\Leftrightarrow-3y^2+6y+9\ge0\)

\(\Leftrightarrow-1\le y\le3\Rightarrow y=\left\{-1;0;1;2;3\right\}\)

Thay lần lượt vào pt ban đầu để tìm x nguyên

ĐKXĐ: ...

\(\Leftrightarrow x^2+\left(x^2+8x\right)+\left(14-2\sqrt{x^2+8x}\right)x-14\sqrt{x^2+8x}+24=0\)

Đặt \(\sqrt{x^2+8x}=a\ge0\) pt trở thành:

\(x^2+a^2+\left(14-2x\right)x-14a+24=0\)

\(\Leftrightarrow x^2-2ax+a^2+14\left(x-a\right)+24=0\)

\(\Leftrightarrow\left(x-a\right)^2+14\left(x-a\right)+24=0\)

\(\Leftrightarrow\left(x-a+2\right)\left(x-a+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x+2\\a=x+12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+8x}=x+2\left(x\ge-2\right)\\\sqrt{x^2+8x}=x+12\left(x\ge-12\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+8x=x^2+4x+4\\x^2+8x=x^2+24x+144\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)

\(P=x^4-8x^3+24x^2-32x+16+3x^2-12x+12-5\)

\(P=\left(x-2\right)^4+3\left(x-2\right)^2-5\ge-5\)

\(\Rightarrow P_{min}=-5\) khi \(x=2\)

trả lời :

P=x⁴ - 8x³ + 27x² - 44x +23

P= (x-2)⁴ + 3(x-2)² - 5 ≥ 5

P_{min= -5 khi x = 2}

các bn tham khảo thôi nha (cs khi sai ráng chịu)

Lời giải:

Ta có:
\(x^4-8x^3+27x^2-44x+23\)

\(=(x^4-8x^3+16x^2)+11x^2-44x+23\)

\(=(x^2-4x)^2+11(x^2-4x)+23\)

\(=(x^2-4x)^2+8(x^2-4x)+16+3(x^2-4x)+7\)

\(=(x^2-4x+4)^2+3(x^2-4x+4)-5\)

\(=(x-2)^4+3(x-2)^2-5\geq -5\)

Vậy GTNN của $P$ là $-5$ khi $x=2$

10:25 x 6,8 và 0,4 x 6,8

10 : 25 x 6,8 = 0,4 x 6,8

                   = 2,72

0,4 x  6,8 = 2,72

Vậy 10 : 25 x 6,8  = 0,4 x  6,8

10 : 8 x 3,2 và 1,25 x 3,2

10 : 8 x 3,2 = 1,25 x 3,2 

                 = 4

1,25 x 3,2 = 4

Vậy 10 : 8 x 3,2 = 1,25 x 3,2

=>4x^2-5x+1=0

=>(x-1)(4x-1)=0

=>x=1 hoặc x=1/4

\(1,6x^2-2x+0,4=0\)

\(\Leftrightarrow\dfrac{8}{5}x^2-2x+\dfrac{2}{5}=0\)

\(\Leftrightarrow\dfrac{8}{5}x^2-\dfrac{8}{5}x-\dfrac{2}{5}x+\dfrac{2}{5}=0\)

\(\Leftrightarrow\dfrac{8}{5}x\left(x-1\right)-\dfrac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{8}{5}x-\dfrac{2}{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{8}{5}x-\dfrac{2}{5}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{8}{5}x=\dfrac{2}{5}\end{matrix}\right.\)

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

Vậy \(S=\left\{1;\dfrac{1}{4}\right\}\)

\(1,6x^2-2x+0,4=0\)

\(\Leftrightarrow2,5\left(1,6x^2-2x+0,4\right)=0\)

\(\Leftrightarrow4x^2-5x+1=0\)

\(\Leftrightarrow4x^2-4x-x+1=0\)

\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\)

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

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......