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\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
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\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
x(2x - 3) - 2(3 - 2x) = 0
x(2x - 3) + 2(2x - 3) = 0
(2x - 3)(x + 2) = 0
\(\left[\begin{array}{nghiempt}2x-3=0\\x+2=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-2\end{array}\right.\)
2x(x - 5) - x(3 + 2x) = 26
2x2 - 10x - 3x - 2x2 = 26
- 13x = 26
x = - 26 : 13
x = - 2
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
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 - 1)3 + 3(x + 1)2 = (x2 - 2x + 4)(x + 2)
x3 - 3x2 + 3x - 1 + 3(x2 + 2x+ 1) = x3 + 8
\(\Rightarrow\)x3 - 3x2 + 3x - 1 + 3x2 + 6x - x3 - 8 = 0
\(\Rightarrow\) 9x - 9 = 0
\(\Rightarrow\) 9x = 9
\(\Rightarrow\) x = 1
b) (2x - 1)(x + 3) - x(3 + 2x) = 26
2x2 + 6x - x - 3 - 3x - 2x2 = 26
2x - 3 = 26
\(\Rightarrow\) 2x = 29
\(\Rightarrow\) x = 14.5
a) ( x - 1)3 + 3(x+1)2 = (x2 - 2x + 4 )( x+ 2)
=>x3-3x2+3x-1+3(x2+2x+1)=x3+8
=>x3-3x2+3x-1+3x2+6x+3-x3-8=0
=>(x3-x3)+(-3x2+3x2)+(3x+6x)+(-1+3-8)=0
=>9x-6=0
=>9x=6
=>x=\(\dfrac{2}{3}\)
mk làm lun nha
a, 2x^2-6x-3x-2x^2=26
-9x=26
x=-26/9
b,x^2+2.x.4+16-(x^2-1)=16
x^2+8x+16-x^2+1=16
8x=-1
x=-1/8
c,(2x)^2-2.2x.1+1-4(x^2-7^2)=0
4x^2-4x+1-4x^2+196=0
-4x=-197
x=197/4
d,x^2-5x-4x+20=0
-9x=-20
x=20/9
**** cho mk nha
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+2^2\right)-x\left(x^2-3^2\right)=26\\ \Leftrightarrow x^3+2^3-x^3+9x=26\\ \Leftrightarrow9x+8=26\\ \Rightarrow x=\dfrac{26-8}{9}=2\)
Lời giải:
$(x+2)(x^2-2x+4)-x(x+3)(x-3)=26$
$\Leftrightarrow x^3+8-x(x^2-9)=26$
$\Leftrightarrow 9x+8=26$
$\Leftrightarrow 9x=18$
$\Leftrightarrow x=2$
\(\left(2x-1\right)^3+3\left(2x+3\right)^2=26\)
\(\Leftrightarrow\left(2x\right)^3-3\left(2x\right)^2+3.2x-1+3\left(4x^2+12x+9\right)=26\)
\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2+36x+27=26\)
\(\Leftrightarrow8x^3+42x+26=26\)
\(\Leftrightarrow8x^3+42x=0\)
\(\Leftrightarrow4x\left(2x^2+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\2x^2+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2=-13\left(L\right)\end{matrix}\right.\)
Vậy \(x=0\)