Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

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

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

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

đúng rồi pn. giúp mik đc bài nào cũng đc

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

                                    \(2x=\frac{3}{5}-\frac{3}{5}\)

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

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

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

ai nhanh tay thì mk sẽ k cho

\(2x^2-7x+5=0\)

\(2x^2-2x-5x+5=0\)

\(2x\left(x-1\right)-5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x-5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)

\(x\left(2x-5\right)-4x+10=0\)

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

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

\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)

\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)

\(x^2-25-x^2+2x=15\)

\(2x=15+25\)

\(2x=40\)

\(x=\frac{40}{2}\)

\(x=20\)

\(x^2\left(2x-3\right)-12+8x=0\)

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

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

\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))

\(2x=3\)

\(x=\frac{3}{2}\)

\(x\left(x-1\right)+5x-5=0\)

\(x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(x+5\right)=0\)

\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)

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

\(4x^2-12x+9-4x^2+4x=5\)

\(-8x=5-9\)

\(-8x=-4\)

\(x=\frac{4}{8}\)

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

\(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)

\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)

\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)

\(\left(2x-5\right)\left(x+11\right)=0\)

\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)

\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)