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a: =>3x+10-2x=0
hay x=-10
c: \(\Leftrightarrow3x^2-3x^2+6x=36\)
=>6x=36
hay x=6
\(a,3x+2\left(5-x\right)=0\)
\(\Rightarrow3x+10-2x=0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
\(b,x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Rightarrow2x^3+9x^2-5x-2x^3-9x^2-4,5=3,5\)
\(\Rightarrow-5x-4,5=3,5\)
\(\Rightarrow-5x=8\)
\(\Rightarrow x=-\dfrac{8}{5}\)
\(c,3x^2-3x\left(x-2\right)=36\)
\(\Rightarrow3x^2-3x^2+6x=36\)
\(\Rightarrow6x=36\)
\(\Rightarrow x=6\)
\(d,\left(3x^2-x+1\right)\left(x-1\right)=x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(\Rightarrow3x^3-3x^2-x^2+x+x-1+4x^2-3x^3=\dfrac{5}{2}\)
\(\Rightarrow2x-1=\dfrac{5}{2}\)
\(\Rightarrow2x=\dfrac{7}{2}\)
\(\Rightarrow x=\dfrac{7}{4}\)
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
b)3x2 - 3x(x - 2)=36 c) (3x2 - x + 1)(x - 1)+ x2(4 - 3x) = 5/2
3x2 - 3x2 + 6x= 36 3x3 - 3x2 - x2 + x + x - 1 + 4x2 - 3x3= 5/2
6x=36 =>x=36 : 6= 6 (3x3 - 3x3) + (-3x2 - x2 + 4x2) + (x + x) - 1= 5/2
2x - 1= 5/2 =>2x= 5/2 + 1= 7/2
x= 7/2 : 2 =7/4
a) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5\)
\(\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}\)
b) \(\left(2x-5\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)
a) x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5x(2x−1)(x+5)−(2x2+1)(x+4,5)=3,5
\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5⇔2x3−x2+10x2−5x−2x3−x−9x2−4,5=3,5
\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}⇔−6x=8⇔x=−34
b) \left(2x-5\right)^2+\left(y-3\right)^2=0(2x−5)2+(y−3)2=0
\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)
1.a) \(\Leftrightarrow\) 3x+10-2x =0
\(\Leftrightarrow\text{ 3x-2x=-10}\)
\(\Leftrightarrow x=-10\)
b) coi lại có thiếu ngoặc ko nhé
cứ nhân vào dấu ngoặc rồi làm như thường
a) 3x + 2( 5 - x) = 0
3x + 10 - 2x = 0
x + 10 = 0
x = -10.
b) x(2x - 1)(x + 5) - ( 2x2 + 1)(x + 4,5) = 3,5
* x(2x2 +10x-x-5) - (2x3 + 9x2 + x + 4,5)=3,5.
2x3 + 10x2 - x2 -5x- 2x3 - 9x2 -x -4,5=3,5
-6x - 4,5 =3,5
-6x = 8
x = -8/6.