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Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
a) \(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)
\(\Leftrightarrow6x-1=5\Leftrightarrow6x=6\Leftrightarrow x=1\)
b) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x+1\right)^2=-10\)
\(\Leftrightarrow\left(x+1-x+1\right)\left[\left(x^2+2x+1+x^2-2x+1+\left(x^2-1\right)\right)\right]-6\left(x^2+2x+1\right)=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6x^2-12x-6=-10\)
\(\Leftrightarrow6x^2+2-6x^2-12x-6=-10\)
\(\Leftrightarrow-12x-4=-10\Leftrightarrow12x=-6\Leftrightarrow x=\dfrac{1}{2}\)
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow-4x+13=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\frac{7}{4}\)
Vậy \(x=1\).
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+37=10\)
\(\Leftrightarrow-24x=27\)
\(\Leftrightarrow x=\frac{9}{8}.\)
Mấy pài kia tương tự . :D
\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
a, (x-2)^2 - (x-3)(x+3)=6
x^2-4x+4-(x^2-9)=6
x^2-4x+4-x^2+9=6
(x^2-x^2)-4x+13=6
-4x=-7
x=1,75
b, 4(x-3)^2 - (2x-1)(2x+1)=10
4(x^2-6x+9)-(4x^2-1)=10
4x^2-24x+36-4x^2+1=10
-24x+37=10
x=9/8
c,(x-4)^2 - (x+2)(x-2)=6
x^2-8x+16-(x^2-4)=6
x^2-8x+16-x^2+4=6
-8x+20=6
x=7/4
d, 9(x+1)^2 - (3x-2)(3x+2)=10
9(x^2+2x+1)-(9x^2-4)=10
9x^2+18x+9-9x^2+4=10
18x+13=10
x=-1/6
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(-4x+13=6\)
\(-4x=6-13\)
\(-4x=-7\)
\(x=\frac{-7}{-4}\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(4x^2-24x+36-4x^2+1=10\)
\(-24x+37=10\)
\(x=\frac{9}{8}\)
Vậy \(x=\frac{9}{8}\)
\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(x^2-8x+16-\left(x^2-4\right)=6\)
\(x^2-8x+16-x^2+4=6\)
\(-8x+20=6\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(9x^2+18x+9-9x^2+4=10\)
\(18x+13=10\)
\(x=\frac{-1}{6}\)
Vậy \(x=\frac{-1}{6}\)
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
giải phóng hết nó ra r` khử từng đứa 1....
cách khác thì dùng HĐT (cho cả 4 phép này) a^2-b^2=(a+b)(a-b)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(x^2-4x+4-x^2+9=6\)
\(-4x=6-4-9\)
\(-4x=-7\)
\(x=\frac{7}{4}\)
\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(x^2-8x+16-\left(x^2-4\right)=6\)
\(x^2-8x+16-x^2+4=6\)
\(-8x=6-16-4\)
\(-8x=-14\)
\(x=\frac{14}{8}=\frac{7}{4}\)