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14 tháng 10 2019
1,\(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)
\(=-6x+5\)
2,\(\left(x+1\right)^2-3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)
\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)
\(=-6x^2+6x+75\)
3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-1\right)^3-\left(x^3-27\right)\)
\(=x^3-3x^2+3x-1-x^3+27\)
\(=-3x^2+3x+26\)
4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)
\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)
\(=x^3+125-x^3-6x^2-12x-8\)
\(=-6x^2-12x+117\)
5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)
\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)
=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)
\(=-x^3+4x^2-4x+1\)
6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)
\(=3x-26\)
7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)
=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)
\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)
\(=-4x^2-27x-58\)
Nếu đúng thì tick cho mk nha ^_^
EC
6 tháng 9 2016
a)\(\left(x-5\right)\left(x+5\right)=\left(x-2\right)\)
\(x^2-25-x+2=0\)
\(x^2-23-x=0\)
\(x.\left(x-1\right)=23\)
Bài này vô lý quá
b)\(\left(3-2x\right)^2-\left(x-5\right)\left(4x+3\right)=2\left(x+5\right)\)
\(9-12x+4x^2-4x^2-3x+20x+15=2x+10\)
\(5x+24=2x+10\)
\(5x+24-2x-10=0\)
\(3x-14=0\)
\(3x=14\)
\(x=\frac{14}{3}\)
Vậy \(x=\frac{14}{3}\)
EC
6 tháng 9 2016
c)\(\left(7-x\right)\left(2x-5\right)-\left(7-x\right)2x=3\left(-5+x\right)\)
\(\left(7-x\right)\left[\left(2x-5\right)-2x\right]=\left(-15\right)+3x\)
\(5x-35=\left(-15\right)+3x\)
\(5x-35+15+3x=0\)
\(8x-20=0\)
\(8x=20\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
HY
28 tháng 2 2020
\(a.DKXD:x\ne5\\ \frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow12x-9=29x-145\\ \Leftrightarrow12x-29x=9-145\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=8\)
Vậy nghiệm của phương trình trên là \(8\)
\(b.DKXD:x\ne\frac{5}{3}\\ \frac{2x-1}{5-3x}=2\\ \Leftrightarrow2x-1=2\left(5-3x\right)\\ \Leftrightarrow2x-1=10-6x\\ \Leftrightarrow2x+6x=1+10\\ \Leftrightarrow8x=11\\ \Leftrightarrow x=\frac{11}{8}\)
Vậy nghiệm của phương trình trên là \(\frac{11}{8}\)
\(c.DKXD:x\ne1\\ \frac{4x-5}{x-1}=2+\frac{x}{x-1}\\ \Leftrightarrow\frac{4x-5}{x-1}=\frac{2\left(x-1\right)+x}{x-1}\\ \Leftrightarrow4x-5=2x-2+x\\ \Leftrightarrow4x-2x-x=5-2\\ \Leftrightarrow x=3\left(tmdk\right)\)
Vậy nghiệm của phương trình trên là \(3\)
\(d.DKXD:x\ne5;-2\\ \frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7x-35=3x+6\\ \Leftrightarrow7x-3x=35+6\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)
Vậy nghiệm của phương trình trên là \(\frac{41}{4}\)
\(e.DKXD:x\ne0;-5\\\Leftrightarrow \frac{2x+5}{2x}-\frac{x}{x+5}=0\\\Leftrightarrow \frac{2x+5}{2x}=\frac{x}{x+5}\\ \Leftrightarrow\left(x+5\right)\left(2x+5\right)=2x.x\\\Leftrightarrow 2x^2+5x+10x+25=2x^2\\ \Leftrightarrow2x^2-2x^2+15x=-25\\ \Leftrightarrow15x=-25\\\Leftrightarrow x=-\frac{5}{3}\)
Vậy nghiệm của phương trình trên là \(-\frac{5}{3}\)
PA
30 tháng 11 2016
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
3 tháng 7 2022
a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
=>1+3x-6=3-x
=>3x-5=3-x
=>4x=8
hay x=2(loại)
b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)
=>8-x-8x+56=-26
=>-9x+64=-26
=>-9x=-90
hay x=10(nhận)
c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)
\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)
=>-7x+10x=12-5
=>3x=7
hay x=7/3(nhận)
TB
0
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=3\)
<=> (x + 5)^2 + (x - 5)^2 = 3(x - 5)(x + 5)
<=> x^2 + 10x + 25 + x^2 - 10x + 25 = 3x^2 - 75
<=> 2x^2 + 50 = 3x^2 - 75
<=> 2x^2 + 50 - 3x^2 = -75
<=> -x^2 + 50 = -75
<=> -x^2 = -75 - 50
<=> -x^2 = -125
<=> x = +-\(\sqrt{125}\)
<=> x = +-5\(\sqrt{5}\)