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

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

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

\(\Rightarrow x^3\left(x-1\right)+5x^2\left(x-1\right)+7x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+5x^2+7x+3\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2+4x+3\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left[x\left(x+3\right)+\left(x+3\right)\right]=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)

\(x-\frac{3}{4}x+\frac{1}{2}>0\)

\(\Leftrightarrow x\left(1-\frac{3}{4}\right)+\frac{1}{2}>0\)

\(\Leftrightarrow\frac{1}{4}x+\frac{1}{2}>0\)

\(\Leftrightarrow\frac{1}{4}x>\frac{1}{2}\)

\(\Leftrightarrow x>2\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Bạn nào có nik ngọc rồng ( vip gà gì cũng đc ) ko chơi nữa cho mình nha , nhớ gửi qua tin nhắn ... XIN ACC NRO Ạ

Ai cho sẽ đc k mỗi ngày

hướng dẫn cách làm-tự làm tiếp nha :)

a) đặt \(k=x^2-4x\), ta có:\(k^2-2k=15\)\(\Rightarrow k^2-2x+1=16\Rightarrow\left(k-1\right)^2=4^2=\left(-4\right)^2\)

b) đặt \(A=x^2-3x\), ta có: \(A^2-2A-8=0\Rightarrow A^2-2A+1=9\Rightarrow\left(A-1\right)^2=3^2=\left(-3\right)^2\)

c)theo đề \(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\x^2-8x+9=0\end{cases}}\)

\(x^2-4x+3=0\Leftrightarrow x^2-4x+4=1\Leftrightarrow\left(x-2\right)^2=1^2=\left(-1\right)^2\)

\(x^2-8x+9=0\Leftrightarrow x^2-8x+16=7\Leftrightarrow\left(x-4\right)^2=\pm\sqrt{7}^2\)

vt ko chi tiết bn ib là đc rùi, sai tớ làm gì T.T 

mà tớ làm mẫu 1 bài thui nha, bài còn lại có cách làm òi. bn tự dựa vô nha

\(\text{Đặt }k=x^2-4x,\text{ta có:}\)

\(\left(x^2-4x\right)^2-2.\left(x^2-4x\right)=15\)

\(\Leftrightarrow k^2-2k=0\)

\(\Leftrightarrow k^2-2k+1=16\)

\(\Leftrightarrow\left(k-1\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}k-1=4\\k-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}k=5\\k=-3\end{cases}}}\)

\(\text{Với }k=5,\text{Ta có: }x^2-4x=5\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x.\left(x-5\right)+\left(x-5\right)=0\Rightarrow\left(x+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

\(\text{Với }k=-3,\text{ta có: }x^2-4x=-3\Rightarrow x^2-4x+3=0\Rightarrow k^2-3x-x+3=0\)

\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\Rightarrow\left(x-1\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy...

b, (x^2-3x)^2- 2(x^2-3x)-8=0

Đặt x^2-3x = k ta có: 

k^2-2k-8=0

<=> (k^2-4k)+(2k-8)=0

<=> k(k-4)+2(k-4)=0

<=> (k-4)(k+2)=0

=> \(\orbr{\begin{cases}k-4=0\\k+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}k=4\\k=-2\end{cases}}\)

Nếu k=4 thì : x^2-3x=4

<=> x^2-3x-4=0

<=>x^2-4x+x-4=0

<=> x(x-4)+(x-4)=0

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

=>\(\orbr{\begin{cases}x-4=0\\x+1=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

Nếu k=-2 thì : x^2-3x=-2

<=> x^2-3x+2=0

<=>x^2-2x-x+2=0

<=> x(x-2)-(x-2)=0

<=> (x-2)(x-1)=0

=> \(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

Vậy x =2 hoặc x=1 hoặc x=4 hoặc x=-1

a)\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\)

     \(4x^2-11x-3-\left(4x^2-29x+7\right)=15\)

     \(4x^2-11x-3-4x^2+29x-7=15\)

      \(18x-10=15\)

       \(x=\frac{25}{18}\)

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

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

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

        \(-4x-4-x+4=0\)

          \(x=0\)

a) \(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x+1\right)=15\)

\(\left(4x+1\right)\left(x-3-x+7\right)=15\)

\(\left(4x+1\right)4=15\)

\(4x+1=15:4\)

\(4x+1=3,75\)

\(x=\frac{11}{16}\)

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

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

\(\left(x+1\right).-4=x-4\)

\(-4x-4-x+4=0\)

\(-5x=0\)

\(x=0\)

k nhé