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NH
11 tháng 11 2020
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
MC
Tìm x
a) (12x-5)(3x-1)-(18x-1)(2x+3)=5
b) (x+2)(x-3)-(x-2)(x+5)=2(x+3)
c) (2x+3)(2x-1)-(2x+5)-(2x-3)=12
0
9 tháng 8 2020
a.\(\left(4x-1\right)-\left(4x+1\right).\left(x-2\right)=12\)
\(\Leftrightarrow4x-1-\left(4x^2-7x-2\right)-12=0\)
\(\Leftrightarrow4x-1-4x^2+7x+2-12=0\)
\(\Leftrightarrow-4x^2+11x-11=0\)
\(\Rightarrow4x^2-11x+11=0\)
\(\Leftrightarrow\left(2x\right)^2-2.2x.\frac{11}{4}+\frac{11^2}{4^2}-\frac{11^2}{4^2}+11=0\)
\(\Leftrightarrow\left(2x-\frac{11}{4}\right)^2+\frac{55}{16}=0\)( VÔ LÝ )
VẬY KHÔNG CÓ GIÁ TRỊ NÀO CỦA x THỎA MÃN PT ĐÃ CHO
b. \(\left(2x-3\right).\left(2x+1\right)-\left(2x-2\right)^2=15\)
\(\Leftrightarrow4x^2-4x-3-4x^2+8x-4-15=0\)
\(\Leftrightarrow4x-22=0\)\
\(\Leftrightarrow x=\frac{11}{2}\)
VẬY PT CÓ NGHIỆM x= 11/2
TT
9 tháng 8 2020
a) \(\left(4x-1\right)-\left(4x+1\right)\left(x-2\right)=12\)
\(\Leftrightarrow4x-1-\left(4x^2-7x-2\right)=12\)
\(\Leftrightarrow4x-1-4x^2+7x+2=12\)
\(\Leftrightarrow4x^2-11x+11=0\)( Pt vô nghiệm )
b) \(\left(2x-3\right)\left(2x+1\right)-\left(2x-2\right)^2=15\)
\(\Leftrightarrow\left(4x^2-4x-3\right)-\left(4x^2-8x+4\right)=15\)
\(\Leftrightarrow4x=22\)
\(\Leftrightarrow x=\frac{11}{2}\)
TN
10 tháng 7 2018
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
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 \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
9 tháng 8 2020
(4x - 1) - (4x + 1)(x - 2) = 12
=> 4x - 1 - 4x2 - 7x - 2 = 12
=> (4x - 7x) + (- 1 - 2) - 4x2 = 12
=> -3x - 3 - 4x2 = 12
=> -3x - 4x2 = 15
=> không tồn tại x
b. (2x - 3)(2x + 1) - (2x - 2)(2x - 2) = 15
=> 2x(2x + 1) - 3(2x + 1) - 2x(2x - 2) + 2(2x - 2) = 15
=> 4x2 + 2x - 6x - 3 - 4x2 + 4x - 4x - 4 = 15
=> (4x2 - 4x2) + (2x - 6x + 4x - 4x) + (-3 - 4) = 15
=> -4x - 7 = 15
=> -4x = 22
=> x = \(-\frac{11}{2}\)
A
9 tháng 8 2020
a, \(\left(4x-1\right)-\left(4x+1\right)\left(x-2\right)=12\)
\(\Leftrightarrow4x-1-4x^2+8x-x+2=12\)
\(\Leftrightarrow11x+1-4x^2=12\)
\(\Leftrightarrow11x-11-4x^2=0\)( vô nghiệm )
b, \(\left(2x-3\right)\left(2x+1\right)-\left(2x-2\right)^2=15\)
\(\Leftrightarrow4x^2+2x-6x-3-4x^2+8x-4=15\)
\(\Leftrightarrow4x-7=15\Leftrightarrow4x=22\Leftrightarrow x=\frac{11}{2}\)
MN
28 tháng 3 2020
bạn làm sai rồi !
\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Leftrightarrow4x+26=-12\)
\(\Leftrightarrow4x=-38\)
\(\Leftrightarrow x=-\frac{19}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)
(2x-1)2+(2x+1)2-2(2x+1)(2x-1)+x=12
=> (2x+1)2-2(2x+1)(2x-1)+(2x-1)2+x=12
=>[(2x+1)2-2(2x+1)(2x-1)+(2x-1)2 ]+x=12
=>[(2x+1)-(2x-1)]2+x=12
=>(2x+1-2x+1)2+x=12
=>22+x=12
=>4+x=12
=>x=8
vậy x=8