Đặt \(\sqrt{5x^2+6x+5}=a,4x=b\left(a\ge0\right)\)

Khi đó Pt

<=> \(a\left(a^2+1\right)=b\left(b^2+1\right)\)

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

MÀ \(a^2+ab+b^2+1>0\)

=> \(a=b\)

=> \(\sqrt{5x^2+6x+5}=4x\)

=> \(\hept{\begin{cases}x\ge0\\11x^2-6x-5=0\end{cases}}\)

=>\(x=1\)

Vậy x=1

5x .(x\(^2\)- + \(\dfrac{1}{5}\)) = 5x\(^3\) - 15x\(^2\)+x

(x-3) (2x-1) =2x\(^2\) - x - 6x + 3 = 2x\(^2\) -7x +3

x\(^2\).(5x\(^3\)- x - \(\dfrac{1}{2}\) ) = 5x\(^5\)- x\(^3\)- \(\dfrac{1}{2}\)x\(^2\)

(x-2) (6x\(^2\) - 5x +1 )= 6x\(^3\)-5x\(^2\)+x -12x\(^2\) + 10x -2 =6x\(^3\) - 17x\(^2\) + 11x -2

Đề có nhầm k

đúng bạn ạ

a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

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

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

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

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)

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

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

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

\(=6x+1+6x-1-2\sqrt{36x^2-1}=12x-2\sqrt{36x^2-1}\)

\(\left(\sqrt{5x-2}-\sqrt{5x+2}\right)^2=5x-2+5x+2-2\sqrt{\left(5x-2\right)\left(5x+2\right)}=10x-2\sqrt{25x^2-4}\)

\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)

\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)

\(=2a^2.2b^2-4a^2b^2=0\)

\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)

\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)

\(=\left[4-11x\right]^2\)

\(=16-88x+121x^2\)

chúc bn học tốt

a) Ta có: \(2x^3+5x^2-3x=0\)

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

\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)

\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!