bài 1: tìm x
a.(x-2)^2-(x-3)(x+3)=6
b. 4(x-3)^2-(2x-1)(2x+1)=10
c. (x-4)^2-(x-2)(x+2)=6
d. 9(x+1)^2-(3x-2)(3x+2)=10
Tìm x , biết : a) (x-2)2- ( x-3 )(x+3) = 6
b) 4(x - 3)2- (2x-1)(2x+1) = 10
c) 4(x-4)2- (x-2)(x+2) = 6
d) 9( x+1 )2 - (3x-2)(3x+2) =10
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
Tìm x biết
a, ( x - 2)2 - (x - 3)(x+3) = 6
b, 4( x - 3)2 --(2x - 1)(2x + 1)= 10
c, ( x - 4)2 - x (x -2(x + 2) = 6
d, 9( x + 1)^2 -(3x -2)(3x+ 2)= 10
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
Tìm x biết
a, (x-2)^2 - (x-3)(x+3)=6
b, 4(x-3)^2 - (2x-1)(2x+1)=10
c,(x-4)^2 - (x+2)(x-2)=6
d, 9(x+1)^2 - (3x-2)(3x+2)=10
giúp mk nha 😉😉😉
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 \(x=\frac{-1}{6}\)
Bài 1 : Tìm thương Q và dư R sao cho A= B.Q+R biết ;
a) A = \(x^4+3x^3+2x^2-x-4\) và B = \(x^2-2x+3\)
b) A = \(2x^3-3x^2+6x-4\) và B = \(x^2-x+3\)
c) A = \(2x^4+x^3+3x^2+4x+9\) và B = \(x^2+1\)
d) A = \(2x^3-11x^2+19x-6\) và B = \(x^2-3x+1\)
c) A= \(2x^4-x^3-x^2-x+1\) và B = \(x^2+1\)
Tìm x , biết :
a) (x-2)2 - ( x-3 )(x+3) = 6
b) 4(x - 3)2 - (2x-1)(2x+1) = 10
c) 4(x-4)2- (x-2)(x+2) = 6
d) 9( x+1 )2 - (3x-2)(3x+2) =10
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) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\Leftrightarrow x=\dfrac{1}{6}\)
Tìm x , biết :
a) (x-2)2 - ( x-3 )(x+3) = 6
b) 4(x - 3)2 - (2x-1)(2x+1) = 10
c) 4(x-4)2 - (x-2)(x+2) = 6
d) 9( x+1 )2 - (3x-2)(3x+2) =10
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
Bài 2: Tìm x, biết: a) (x+2)(x² -2x+4)-x(x²+2)=15 b) (x-2)³-(x-4)(x² + 4x+16) + 6(x+1)=49 c) (x - 1)³ + (2 - x)(4 + 2x + x²)+ 3x(x + 2) = 16 d) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 15
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 \(x=-\dfrac{13}{24}\)
1.Tìm x
a, (x-2)^2-(x-3)(x+3)=6
b, 4 (x-3)^2-(2x-1)(2x+1)=10
c,x^2-16-3 (x+4)=0
d,(x-4)^2-(x-2)(x+2)=6
e, 9 (x+1)^2-(3x-2 )(3x+2)=10
2.Tìm GTLN
B=-2x^2+6x-11
Các bạn giúp mình với , mình sắp thi r
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=-\(\frac{2}{3}\)
d,5x+10/4x-8.4-2x/x+2
Bài 2: rút gọn
a, 6x ² y ³/8x ³y ²
b, x ³-x/3x+3
c, x ²+3xy/x ²-9y ²
d, x ²+4x+4/3x+6
Bài 3: Thực hiện phép tính
a, (x/x-3+(9-6x/x ²-3x)
b, 1/x-1/x+1
c, (x-12/6x-36)+(6/x ²-6x)
d, (6x-3/x):(4x ²-1/3x ²)
e, (x+y/2x-2y)-(x-y/2x+2y)-(y ²+x ²/y ²-x ²)
f, 7x+6/2x(x+7)-3x+6/2x ²+14x
g, (2/x+2-4/x ²+4x+4):(2/x ²-4+1/2-x)