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1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
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
^^
a) x(2x-7)-4x+14=0
=>x(2x-7)-2(2x-7)=0
=>(x-2)(2x-7)=0
=>x-2=0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
b, x(x-1)+2x-2=0
=>x(x-1)+2(x-1)=0
=>(x+2)(x-1)=0
=>x+2=0 hoặc x-1=0
=>x=-2 hoặc x=1
c, 2x^3+3x^2+2x+3=0
=>x2(2x+3)+2x+3=0
=>(x2+1)(2x+3)=0
=>x2+1=0 hoặc 2x+3=0
Vì x2+1>0 với mọi x ->vô nghiệm
=>2x+3=0 =>x=-3/2
d, x^3+6x^2+11x+6=0
=>x3+3x3+2x+3x2+3x3+6=0
=>x(x2+3x+2)+3(x2+3x+2)=0
=>(x2+3x+2)(x+3)=0
=>[x2+x+2x+2](x+3)=0
=>[x(x+1)+2(x+1)](x+3)=0
=>(x+1)(x+2)(x+3)=0
=>x+1=0 hoặc x+2=0 hoặc x+3=0
=>x=-1 hoặc x=-2 hoặc x=-3
a) x(2x-7)-4x+14=0
=>x(2x-7)-2(2x-7)=0
=>(x-2)(2x-7)=0
=>x-2=0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
b, x(x-1)+2x-2=0
=>x(x-1)+2(x-1)=0
=>(x+2)(x-1)=0
=>x+2=0 hoặc x-1=0
=>x=-2 hoặc x=1
c, 2x^3+3x^2+2x+3=0
=>x2(2x+3)+2x+3=0
=>(x2+1)(2x+3)=0
=>x2+1=0 hoặc 2x+3=0
Vì x2+1>0 với mọi x ->vô nghiệm
=>2x+3=0 =>x=-3/2
d, x^3+6x^2+11x+6=0
=>x3+3x3+2x+3x2+3x3+6=0
=>x(x2+3x+2)+3(x2+3x+2)=0
=>(x2+3x+2)(x+3)=0
=>[x2+x+2x+2](x+3)=0
=>[x(x+1)+2(x+1)](x+3)=0
=>(x+1)(x+2)(x+3)=0
=>x+1=0 hoặc x+2=0 hoặc x+3=0
=>x=-1 hoặc x=-2 hoặc x=-3
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
Bài 1:
a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)
\(=12x^2-4x-6x-2-x-3\)
\(=12x^2-11x-5\)
b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)
\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)
\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)
\(=2x^4y+1x^3y^2-2x^2y^3\)
c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)
\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)
\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)
\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)
\(=-48x^3-8x^2-24x^4+15x-9\)
2) \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)
bn giải tiếp nha
3) \(x^3-4x^2+x+6=0\)
\(\Leftrightarrow\)\(x^3-3x^2-x^2+3x-2x+6=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-x-2\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)
lm tiếp nha
4) \(x^3-3x^2+4=0\)
\(\Leftrightarrow\)\(x^3+x^2-4x^2-4x+4x+4=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\)\( \left(x+1\right)\left(x-2\right)^2=0\)
lm tiếp nha
Mk làm mẫu 1 bài cho nha !
1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0
<=> (x-1).(x^2+5x+6) = 0
<=> (x-1).[(x^2+2x)+(3x+6)] = 0
<=> (x-1).(x+2).(x+3) = 0
<=> x-1=0 hoặc x+2=0 hoặc x+3=0
<=> x=1 hoặc x=-2 hoặc x=-3
Vậy ..............
Tk mk nha