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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) <=> \(2x^2-8x+3x-12+x^2-7x+10=3x^2-5x-12x+20\)
<=> \(2x^2-8x+3x-12+x^2-7x+10-3x^2+5x+12x-20=0\)
<=> \(5x-22=0\)
<=> \(5x=22\)
<=> \(x=\frac{22}{5}\)
b) <=> \(24x^2-9x+16x-6-4x^2-7x-16x-28=10x^2+5x-2x-1\)
<=> \(24x^2-9x+16x-6-4x^2-7x-16x-28-10x^2-5x+2x+1=0\)
<=> \(10x^2-19x-33=0\)
<=> \(10x^2-30x+11x-33=0\)
<=> \(10x\left(x-3\right)+11\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(10x+11\right)=0\)
<=> \(x=3;x=-\frac{11}{10}\)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)-\left(3x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(2x^2-8x+3x-12\right)+\left(x^2-2x-5x+10\right)-\left(3x^2-12x-5x+20\right)=0\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20=0\)
\(\Leftrightarrow5x-22=0\)
\(\Leftrightarrow5x=22\)
\(\Leftrightarrow x=\dfrac{22}{5}\)
Vậy \(x=\dfrac{22}{5}\)
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) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
(2x+3)(x-4)+(x-5)(x-2)-(3x-5)(x-4)=0\(2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20\)=0
(\(2x^2+x^2-3x^2\))+(-8x+3x-2x-5x+12x+5x)+(-12+10-20) =0 5x-22 =0
5x = 22
x = \(\dfrac{22}{5}\)
Vậy x= \(\dfrac{22}{5}\)
b) (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)
(8x-3)(3x+2)-(4x+7)(x+4)-(2x+1)(5x-1)=0
\(24x^2\) +16x-9x-6\(-4x^2\) -16x-7x-28\(-10x^2\) +2x-5x+1=0
(24\(x^2-4x^2-10x^2\))+(16x-9x-16x-7x+2x-5x)+(-6-28+1)=0
10\(x^2-19x-33\)=0
10\(x^2+11x-30x-33=0\)
x(10x+11)-3(10x+11)=0
(x-3) (10x+11)=0
=>x-3=0 => x=3 =>x=3
10x+11=0 10x=-11 x=\(\dfrac{-11}{10}\)
Vậy x=3 hoặc x=\(\dfrac{-11}{10}\)
Tìm x, biết:
a) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
<=> \(2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
<=> \(2x^2-8x+3x+x^2-2x-5x-3x^2+12x+5x=12-10+20\)
<=> \(5x=22\)
<=> \(x=\dfrac{22}{5}\)
Vậy \(S=\left\{\dfrac{22}{5}\right\}\)
a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+16x-9x-4x^2-16x-7x-10x^2+2x-5x=6+28-1\)
\(\Leftrightarrow10x^2-19x=33\)
\(\Leftrightarrow10x^2-19x+33=0\)
Phương trình vô nghiệm!!!!!!!!
b) \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)
\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)
\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)