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A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)
B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)
C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)
D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)
a)(x+2)(x+3)-(x-2)(x+5)=0
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
<=>2x=-16
<=>x=-8
b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)
c)(8-4x)(x+2)+4(x-2)(x+1)=0
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
d)(2x-3)(8x+2)=(4x+1)(4x-1)-3
\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)
\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
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. \(2.\left(5x-8\right)-3.\left(4x-5\right)=4.\left(3x-4\right)+11\Leftrightarrow10x-16-12x+15=12x-16+11\\ \)
\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow14x-4=0\Leftrightarrow x=\frac{2}{7}\)
\(a,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=-16+11+16-15\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\frac{-4}{-14}=\frac{2}{7}\)
\(\Leftrightarrow-2x+1-x-2=8\cdot\left(-4x^2+6x-2x\right)+4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow-3x-1+32x^2-48x+16x-4x^2+8x-4=0\)
\(\Leftrightarrow28x^2-27x-5=0\)
\(\text{Δ}=\left(-27\right)^2-4\cdot28\cdot\left(-5\right)=1289>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{27-\sqrt{1289}}{56}\\x_2=\dfrac{27+\sqrt{1289}}{56}\end{matrix}\right.\)
a) \(\left(x+6\right)^2-x\left(x+9\right)=0\)
\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)
\(\Leftrightarrow\)\(3x+36=0\)
\(\Leftrightarrow\)\(x=-12\)
Vậy...
b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)
\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)
\(\Leftrightarrow\)\(23x+12=9\)
\(\Leftrightarrow\)\(x=-\frac{3}{23}\)
Vậy
c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)
\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)
\(\Leftrightarrow\)\(16x^2+2x-14=0\)
\(\Leftrightarrow\)\(8x^2+x-7=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)
Vậy
d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)
\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)
\(\Leftrightarrow\)\(-12x+16=0\)
\(\Leftrightarrow\)\(x=\frac{4}{3}\)
Vậy
e) \(\left(x-2\right)^2-\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x^2-4x+4-2x^2+x+6=0\)
\(\Leftrightarrow\)\(-x^2-3x+10=0\)
\(\Leftrightarrow\)\(\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
Vậy
\(a,3x\left(x-4\right)-2x+8=0\)
\(\Rightarrow3x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\4\end{cases}}}\)
Vậy \(x=\frac{2}{3}\)hoặc \(x=4\)
\(b,\left(3x-1\right)^2-\left(3x+2\right)\left(3x+1\right)=2\)
\(\Rightarrow9x^2-6x+1-\left(9x^2+3x+6x+1\right)-2=0\)
\(\Rightarrow9x^2-6x+1-9x^2-3x-6x-1-2=0\)
\(\Rightarrow-15x-2=0\)
\(\Rightarrow-15x=2\)
\(\Rightarrow x=\frac{-2}{15}\)
a/ \(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\frac{22}{5}\)
b/ \(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c/ \(\Leftrightarrow24x^2+7x-6-4x^2-9x+28=10x^2+3x-1-33\)
\(\Leftrightarrow10x^2-5x+56=0\)
Phương trình vô nghiệm (chắc do bạn ghi sai đề)
a/ ⇔2x2−5x−12+x2−7x+10=3x2−17x+20⇔2x2−5x−12+x2−7x+10=3x2−17x+20
⇔5x=22⇔5x=22
⇒x=225⇒x=225
b/ ⇔−5x2−2x+16+4x2−4x−8+2x2−8=0⇔−5x2−2x+16+4x2−4x−8+2x2−8=0
⇔x2−6x=0⇔x2−6x=0
⇔x(x−6)=0⇒[x=0x=6⇔x(x−6)=0⇒[x=0x=6
c/ ⇔24x2+7x−6−4x2−9x+28=10x2+3x−1−33⇔24x2+7x−6−4x2−9x+28=10x2+3x−1−33
⇔10x2−5x+56=0⇔10x2−5x+56=0
Phương trình vô nghiệm (chắc do bạn ghi sai đề)
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1