c.

ĐLXĐ: \(x\ge-\dfrac{1}{3}\)

\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)

Đặt \(\sqrt{3x+1}=t\ge0\)

\(\Rightarrow-t^2+t+4x^2-10x+6=0\)

\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{61}{12}\)

\(\Leftrightarrow36x^2+12x-58-2\sqrt{12x+61}=0\)

\(\Leftrightarrow\left(36x^2+24x+4\right)-\left(12x+61+2\sqrt{12x+61}+1\right)=0\)

\(\Leftrightarrow\left(6x+2\right)^2-\left(\sqrt{12x+61}+1\right)^2=0\)

\(\Leftrightarrow\left(6x+1-\sqrt{12x+61}\right)\left(6x+3+\sqrt{12x+61}\right)=0\)

\(\Leftrightarrow...\) tương tự câu a

a.

ĐKXĐ: \(x\ge-\dfrac{5}{4}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

Bài này đơn giản thôi. 

Đặt f(x) =  6x⁴ - 18x³ + 23x² - 13x + 4 > 0

\(f\left(x\right)=\frac{47}{54}+\frac{1}{54}\left(18x^2-27x+13\right)^2+\frac{5}{6}x^2\)

Thao tác trên Maple (vào thống kê hỏi đáp xem ảnh)

Còn cách phân tích bằng tay thì qua VMF có bài viết của mình nói về điều này nhé.

mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk

\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

câu thứ 2 bạn ktra lại đề

\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

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

\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

a)  \(x^5-7x^4-x^3+43x^2-36\)

\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)

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

c)  \(x^4+2x^3-15x^2-18x+64\)

\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

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

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

e)  \(x^3-3x^2-4x+12=x\left(x^2-4\right)-3\left(x^2-4\right)\)

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

a) Ta có: (x-2)(y+1)=23

⇔x-2;y+1∈Ư(23)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=23\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+1=-23\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=23\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-23\\y+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=22\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-24\end{matrix}\right.\\\left\{{}\begin{matrix}x=25\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=-21\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Vậy: (x,y)∈{(3;22);(1;-24);(25;0);(-21;-2)}

c) Ta có: \(-15x-18x+13x=-4600\)

\(\Leftrightarrow-20x=-4600\)

hay x=230

Vậy: x=230