<=>x^2-4x+4=x^2+8x+16

<=>12x+12=0

<=>x=-1

vcs (x-7_là j thế sửa lại đi

x=-1 hoặc 12

<=>(x-4)(x-5)(x-6)(x-7)-1680=(x-12)(x+1)(x^2-11x+70)

=>x-12=0

=>x=12

=>x+1=0

=>x=-1

áp dụng denta ta có : 

x^2-11x+70=0

\(\Rightarrow\left(-11\right)^2-4\left(1.70\right)=-159\)

=>D<0 ko biết viêt hình tam giác nên thay = D( ko có nghiệm thực )

=>x=-1 hoặc 12

đoạn cuối: \(x\left(x-11\right)=12\Leftrightarrow x^2-11x=12\)

\(\Rightarrow x^2-11x+\)\(\dfrac{121}{4}=\dfrac{169}{4}\)

\(\Rightarrow\left(x^2-\dfrac{11}{2}\right)^2-\dfrac{169}{4}=0\)

\(\Rightarrow\left(x^2-\dfrac{11}{2}+\dfrac{13}{2}\right)\left(x^2-\dfrac{11}{2}-\dfrac{13}{2}\right)=0\)

\(\Rightarrow\left(x^2+1\right)\left(x^2-12\right)=0\)

\(\Rightarrow x^2-12=0\Leftrightarrow x=\pm\sqrt{12}\)

=>(x-4)(x-7)(x-5)(x-6)=1680

(x^2-11x+28)(x^2-11x+30)=1680

(x^2-11x+29-1)(x^2-11x+29+1)=1680

(x^2-11x+29)^2-1^2=1680

(x^2-11x+29)^2=1680+1=1681

<=>x^2-11x+29=41

x(x-11)=41-29=12

xet cac uoc cua 12 roi thay vao x , neu thoa man thi chinh la nghiem cua phuong trinh

Ta có : \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)

\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)-1680=0\)

Đặt \(t=\left(x^2-11x+28\right)\)

\(\Rightarrow t\left(t+2\right)-1680=0\)

\(\Rightarrow t^2+2t+1-1681=0\)

\(\Rightarrow\left(t+1\right)^2-41^2=0\)

\(\Rightarrow\left(t-40\right)\left(t-42\right)=0\)

\(\Rightarrow\left(x^2-11x-12\right)\left(x^2-11x-12\right)=0\)

\(\Rightarrow x\left(x-11\right)=12\)

Đến đây bạn tự giải nha

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)

\(\left(x+5\right)+\left(x-5\right)+5x+x\div5=180\)

\(\Leftrightarrow\left(x+x+5x\right)+\left(5-5\right)+\frac{x}{5}=180\)

\(\Leftrightarrow7x+0+\frac{x}{5}=180\)

\(\Leftrightarrow7x+\frac{x}{5}=180\)

\(\Leftrightarrow\frac{35x+x}{5}=180\)

\(\Leftrightarrow35x+x=180.5\)

\(\Leftrightarrow36x=900\)

\(\Leftrightarrow x=\frac{900}{36}\)

\(\Leftrightarrow x=25\)

Vậy phương trình có 1 nghiệm duy nhất là 25

(x + 5) + (x - 5) + 5x + \(\frac{x}{5}\)= 180

<=> x + 5 + x - 5 + 5x + \(\frac{x}{5}\) = 180

<=> 7x + \(\frac{x}{5}\) = 180

<=> \(\frac{36x}{5}=180\)

\(\Leftrightarrow x=\frac{180.5}{36}=25\)

\(\left(x+5\right)+\left(x-5\right)+\left(x.5\right)+\left(x:5\right)=180\)\(\Leftrightarrow2x+5x+\frac{x}{5}=180\Leftrightarrow7x+\frac{x}{5}=180\)

\(\Leftrightarrow\frac{35x+x}{5}=\frac{900}{5}\Leftrightarrow35x+x=900\Leftrightarrow36x=900\Leftrightarrow x=25\)

Vậy phương trình có tập nghiệm S = { 25 }

\(\dfrac{x+2}{2016}+\dfrac{x+3}{2015}+\dfrac{x+4}{2014}+\dfrac{x+2036}{6}=0\)

<=>\(\dfrac{x+2}{2016}+1+\dfrac{x+3}{2015}+1+\dfrac{x+4}{2014}+1+\dfrac{x+2036}{6}-3=0\)

<=>\(\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}+\dfrac{x+2018}{2014}+\dfrac{x+2018}{6}=0\)

<=>\(\left(x+2018\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{6}\right)=0\)

vì 1/2016+1/2015+1/2014+1/6 khác 0

=>x+2018=0<=>x=-2018

vậy...................

chúc bạn học tốt ^ ^