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a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a)\(x^2+x-x^2+2=0\)\(\Rightarrow x+2=0\)\(\Rightarrow x=-2\)
b)\(2\left(3x+2\right)-2\left(x+6\right)=0\)
\(\Rightarrow2\left(3x+2-x-6\right)=0\)
\(\Rightarrow2\left(2x-4\right)=0\)
\(\Rightarrow2x-4=0\Rightarrow x=2\)
c)\(4x^4-6x^3-4x^4+6x^3-2x^2=0\)
\(\Rightarrow-2x^2=0\Rightarrow x=0\)
d)\(\left(3x^2-x-2\right)-3\left(x^2-x-2\right)=4\)
\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)
\(\Rightarrow2x+4=4\Rightarrow2x=0\Rightarrow x=0\)
\(a,\left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x=5\\ \Rightarrow-2x=-4\\ \Rightarrow x=2\\ b,x\left(x-6\right)+2x-12=0\\ \Rightarrow x\left(x-6\right)+2\left(x-6\right)=0\\ \Rightarrow\left(x+2\right)\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)
c nhường mấy bn khácccc
a) x^2-6x+9=4.
x=1, x=5
b) x^2(x-3)-(4X-12)=0
x=-2, x=2, x=3
c) (2x+3)^2-4(x+2)^2=12
x=-19/4
a) (x2+2x+1)+(y2+2y+1)=0
=>(x+1)2+(y+1)2=0
Vì\(\left(x+1\right)^2\ge0;\left(y+1\right)^2\ge0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
Vậy x=y=-1
Bạn làm tiếp câu còn lại nha <3
Chúc bạn học tốt :)
b: \(3x^2-2x-1=0\)
=>\(3x^2-3x+x-1=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: Bạn ghi lại đề đi bạn
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
2x^2+12+2x(4-x)=0
2x^2+12+8x-2x^2=0
8x+12=0
8x=-12
x=-12/8=-3/2