a) \(4x-\sqrt[]{3\left(3x-1\right)}=3x-1\)

\(\Leftrightarrow\sqrt[]{3\left(3x-1\right)}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3\left(3x-1\right)=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\9x-3=x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\left(a\right)\\x^2-7x+4=0\left(1\right)\end{matrix}\right.\)

Giải \(pt\left(1\right):\)

\(\Delta=49-16=33\Rightarrow\sqrt[]{\Delta}=\sqrt[]{33}\)

Phương trình (1) có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt[]{33}}{2}\\x=\dfrac{7-\sqrt[]{33}}{2}\end{matrix}\right.\) (thỏa \(\left(a\right)\))

câu 2: c,4x^4 y^4

\(2,B\\ 3,D\\ 4,D\\ 5,B,C\\ 6,A\\ 7,C\\ 8,A\)

de qua

x.(2.x-1)+1/3-2/3.x=0

Bài 2 :

\(A=4x^2-2.2x.2+4+1\)

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

Thấy : \(\left(2x-2\right)^2\ge0\)

\(A=\left(2x-2\right)^2+1\ge1\)

Vậy \(MinA=1\Leftrightarrow x=1\)

\(B=\left(5x\right)^2-2.5x.1+1-4\)

\(=\left(5x-1\right)^2-4\)

Thấy : \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)

Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)

\(C=\left(7x\right)^2-2.7x.2+4-5\)

\(=\left(7x-2\right)^2-5\)

Thấy : \(\left(7x-2\right)^2\ge0\)

\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)

Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)

\(1.\)

\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)

\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)

\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5

\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)

\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)

\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)

dấu"=" xảy ra<=>x=1/4

\(a,x^3+3x^2=4x+12\)

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

\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)

\(b,49x^2=\left(3x+2\right)^2\)

\(7x=3x+2\)

\(\Rightarrow7x-3x=2\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\frac{1}{2}\)

các câu còn lại tương tự nha

\(a,x^3+3x^2=4x+12\)

\(x^3+3x^2-4x-12=0\)

\(\Rightarrow x^2\left(x+3\right)-4\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\\left(x+2\right)\left(x-2\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)

\(b,49x^2=\left(3x+2\right)^2\)

\(\Rightarrow\left(7x\right)^2=\left(3x+2\right)^2\)

\(\Rightarrow7x=3x+2\)

\(\Rightarrow7x-3x=2\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\frac{1}{2}\)

\(c,3x^2\left(x-5\right)+12\left(5-x\right)=0\)

\(3x^2\left(x-5\right)-12\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x^2-12\right)=0\)

\(\Rightarrow3.\left(x-5\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}}\)

\(d,x^2\left(x-5\right)+45-9x=0\)

\(x^2\left(x-5\right)+9\left(5-x\right)=0\)

\(x^2\left(x-5\right)-9\left(x-5\right)=0\)

\(\left(x-5\right)\left(x^2-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)

\(A-B-C\)

\(=\left(3x^4-2x^3-x+1\right)-\left(-2x^3+4x^2+5x\right)-\left(-3x^4+2x^2+5\right)\)

\(=3x^4-2x^3-x+1+2x^3-4x^2-5x+3x^4-2x^2-5\)

\(=6x^4-6x^2-6x-4\)

a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)

b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

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