tìm x biết
a,(x-2)^2-(x+3)^2-4(x+1)=5
b, (2x-3)(2x+3)-(x-1)^2-3x(x-5)=-44
c, (5x+1)^2-(5x+3)(5x-3)=30
d,(x+3)^2+(x-2)(x+2)-2(x-1)^2=7
giúp mik vs mik cần gấp
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\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
a) Cậu xem lại đề đi
b) \(3x.\left(x-2\right)-5x.\left(1-x\right)-8.\left(x^2-3\right)=4\)\(\Leftrightarrow3x^2-6x-5x+5x^2-8x^2+24-4=0\Leftrightarrow-11x+20=0\Leftrightarrow-11x=-20\Leftrightarrow x=\frac{20}{11}\)
c) \(2x^2+3.\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\Leftrightarrow2x^2+3\left(x^2-1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\Leftrightarrow-5x=3\Leftrightarrow x=-\frac{3}{5}\)
Trần Anh: Cảm ơn bạn nhiều nhé :)) Phần a đúng là có sai đề pạn ạ mik làm hoài mà cux ko ra hì hì !!~~ Dù sao mik cux cảm ơn pạn nhiều nhiều nhé :3
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
Tìm x
a) Ta có: \(3\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow3\left(-4x^2+5x-1\right)+4\left(3x^2+11x+6\right)-38=0\)
\(\Leftrightarrow-12x^2+15x-3+12x^2+44x+24-38=0\)
\(\Leftrightarrow59x-17=0\)
\(\Leftrightarrow59x=17\)
hay \(x=\frac{17}{59}\)
Vậy: \(x=\frac{17}{59}\)
b) Ta có: \(5\left(2x+3\right)\left(x+2\right)-2\left(5x-4\right)\left(x-1\right)=75\)
\(\Leftrightarrow5\left(2x^2+4x+3x+6\right)-2\left(5x^2-5x-4x+4\right)-75=0\)
\(\Leftrightarrow5\left(2x^2+7x+6\right)-2\left(5x^2-9x+4\right)-75=0\)
\(\Leftrightarrow10x^2+35x+30-10x^2+18x-8-75=0\)
\(\Leftrightarrow53x-53=0\)
\(\Leftrightarrow53x=53\)
hay x=1
Vậy: x=1
c) Ta có: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Leftrightarrow5x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)
\(\Leftrightarrow-5x=-3\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
d) Ta có: \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2-4\right)=0\)
\(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{0;6\right\}\)