a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)

\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)

Trả lời :.....................................

\(\Leftrightarrow24x=27\Leftrightarrow x=1,125..........................\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A

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

cậu khai triển các tích ra là ra thui mà cậu

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 \(x=\frac{-1}{6}\)

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

b: =>4(x^2-6x+9)-4x^2+1=10

=>4x^2-24x+36-4x^2+1=10

=>-24x+37=10

=>-24x=-27

=>x=9/8

c: =>4x^2-32x+64-x^2+4=6

=>3x^2-32x+62=0

hay \(x=\dfrac{16\pm\sqrt{70}}{3}\)

d: =>9x^2+18x+9-9x^2+4=10

=>18x+23=10

=>18x=-13

=>x=-13/18

Định làm mà mệt quá

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

2,

B=-2x²+6x-11=-2(x²-3x+\(\frac{4}{9}\)) - \(\frac{91}{9}\)

=-2(x+\(\frac{2}{3}\))²-\(\frac{91}{9}\)

vì (x+\(\frac{2}{3}\))² \(\ge\)0  với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))²\(\le\)0 với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))² -\(\frac{91}{9}\)\(\le-\frac{91}{9}\)với mọi x\(\in\)R

dấu bằng xảy ra khi -2(x+\(\frac{2}{3}\))² =0 hay x+\(\frac{2}{3}\)=0 =>x=-\(\frac{2}{3}\)

vậy GTLN của B=-\(\frac{91}{9}\)khi x=-\(\frac{2}{3}\)

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 \(x=-\dfrac{13}{24}\)

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 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)