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\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Rightarrow x^2-4x+4-x^2+9=6\)
\(\Rightarrow-4x+13=6\)
\(\Rightarrow x=\dfrac{6-13}{-4}=\dfrac{7}{4}\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+1=10\)
\(\Rightarrow4x^2-24x+36-4x^2+1=10\)
\(\Rightarrow-24x+37=10\)
\(\Rightarrow x=\dfrac{10-37}{-24}=\dfrac{27}{24}\)
\(c,x^2-16-3\left(x+4\right)=0\)
\(\Rightarrow x^2-16-3x-12=0\)
\(\Rightarrow x^2-3x-28=0\)
\(\Rightarrow x^2-7x+4x-28=0\)
\(\Rightarrow\left(x-7\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
\(d,\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Rightarrow x^2-8x+16-x^2+4=6\)
\(\Rightarrow-8x+20=6\)
\(\Rightarrow x=\dfrac{6-20}{-8}=\dfrac{-14}{-8}=\dfrac{7}{4}\)
\(e,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Rightarrow9x^2+18x+9-9x^2+4=10\)
\(\Rightarrow18x+13=10\)
\(\Rightarrow x=\dfrac{10-13}{18}=\dfrac{-3}{18}=\dfrac{-1}{6}\)
a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)
\(\Leftrightarrow3x+2x-10=6-5x+1\)
\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm
b, \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)
Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }
c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3+x^2-3x-2=0\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn
Vậy ...
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
bạn đăng tách ra nhé
a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)
\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)
b, sửa đề : \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)
\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)
c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)
d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)
\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)
e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)
\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)
f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)
\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)
g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)
\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)
vô nghiệm
a) (x-2)2 -(x-3)(x-3)=6
=>x2 -4x+4-x2+3=6
=>7-4x=6
=>4x=1 =>x=\(\frac{1}{4}\)
b)4(x-3)2 -(2x-1)(2x+1)=10
=>4(x2-6x+9)-4x2+1=10
=>4x2-24x+36-4x2+1=10
=>37-24x=10 =>24x=27 =>x=\(\frac{9}{8}\)
c)x2-16-3(x+4)=0
=>(x-4)(x+4)-3(x+4)=0
=>(x-7)(x+4)=0
=>\(\orbr{\begin{cases}x-7=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-4\end{cases}}}\)
=>x\(\in\left\{-4;7\right\}\)
d)(x-4)2-(x-2)(x+2)=6
=>x2-8x+16-x2+4=6
=>20-8x=6
=>8x=14 =>x=\(\frac{4}{7}\)
e) 9(x+1)2-(3x-2)(3x+2)=10
=>9(x2 +2x+1)-9x2+4=10
=>9x2+18x+9-9x2+4=10
=>18x+13=10
=>18x=-3
=>x=\(\frac{-1}{6}\)
mình chỉ làm bài 1 nha
nhớ chon mk đúng nha
Cảm ơn bạn nha . Ai giúp mình làm bài 2 với TT