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![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=\frac{-1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\Leftrightarrow x=\dfrac{7}{4}\)
b) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\Leftrightarrow x=\dfrac{9}{8}\)
c) đề có sai ko vậy bạn :
\(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=6\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\Leftrightarrow x=\dfrac{7}{4}\)
d) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\Leftrightarrow x=\dfrac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
a) ( x - 2 )2 - ( x - 3 )( x+ 3) = 6
( x2 - 4x + 4 ) - ( x2 - 9 ) = 6
x2 - 4x + 4 - x2 + 9 = 6
-4x + 13 = 6
-4x = -7
x = 7/4
bạn làm giúp mình caau c và câu d nữa với . cảm ơn