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a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)
b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)
d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)
a) 4x(3x-7)-6(2x2-5x+1)=12
=>4x.3x-4x.7-6.2x2-6.(-5x)-6.1=12
=>12x2-28x-12x2+30x-6=12
=>2x-6 =12
=>2x =12+6
=>2x =18
=>x =18:2
=>x =6
b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27
=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27
=>20x2-5x+12x-3-20x2+30x+14x-21=27
=>39x-36 =27
=>39x =27+36
=>39x =63
=>x =63:39
=>x =21/13
c) (8x-5)(3x+2)-(12x+7)(2x-1)=17
=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17
=>24x2+16x-15x-10-24x2+12x+14x-7=17
=>27x-17 =17
=>27x =17+17
=>27x =34
=>x =34:27
=>x =34/27
d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190
=>30x2-5x+63x-9 - 30x2-2x-45x-3=-190
=>11x-12 =-190
=>11x =-190+12
=>11x =-178
=>x = -178:11
=>x =-178/11
a) (5x3 - 3x2) : x2 = 7
5x3 : x2 - 3x2 : x2 = 7
5x - 3 = 7
5x = 7 + 3
5x = 10
x = 10 : 5
x = 2
Vậy x = 2
b) (12x3 - 8x) : x - 4x(3x - 1) = 0
12x3 : x - 8x : x - 12x2 + 4x = 0
12x2 - 8 - 12x2 + 4x = 0
-8 + 4x = 0
4x = 0 + 8
4x = 8
x = 8 : 2
x = 4
Vậy x = 4
a) 4x - 10 - x^2
= - ( x^2 - 4x + 10)
= - ( x^2 - 4x + 4 + 6 )
= - ( x- 2 )^2 - 6
Vì -( x - 2 )^2 <=0 => - ( x- 2 )^2 - 6 <0
VẬy GTBT luôn âm
Tương tự
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy \(x\in\left\{3;-\frac{11}{10}\right\}.\)
Bài làm :
Ta có :
\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Leftrightarrow10x^2-30x+11x-33=0\)
\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
Vậy x=3 hoặc x=-11/10
Tìm GTLN:
\(A=-x^2+6x-15\)
\(=-\left(x^2-6x+15\right)\)
\(=-\left(x^2-2.x.3+9+6\right)\)
\(=-\left(x+3\right)^2-6\le0\forall x\)
Dấu = xảy ra khi:
\(x-3=0\Leftrightarrow x=3\)
Vậy Amax = - 6 tại x = 3
Tìm GTNN :
\(A=x^2-4x+7\)
\(=x^2+2.x.2+4+3\)
\(=\left(x+2\right)^2+3\ge0\forall x\)
Dấu = xảy ra khi:
\(x+2=0\Leftrightarrow x=-2\)
Vậy Amin = 3 tại x = - 2
Các câu còn lại làm tương tự nhé... :)
a,4x^2-4x+1=0
4x^2-2x-2x+1=0
2x (2x-1)-(2x-1)=0
(2x-1)(2x-1)=0
(2x-1)^2=0
=>2x-1=0 <=> x=1/2
( 4x - 1 )3 + ( 3 - 4x )( 9 + 12x + 16x2 ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )
<=> 64x3 - 48x2 + 12x - 1 + [ 33 - ( 4x )3 ] = ( 8x )2 - 1 - 3x + 5
<=> 64x3 - 48x2 + 12x - 1 + 27 - 64x3 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 - 64x2 + 3x - 4 = 0
<=> -112x2 + 15x + 22 = 0 (*)
\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)
Lớp 8 sao nghiệm xấu thế -..-
\(\Leftrightarrow3x-4+5x-7=15\)
=>8x=26
hay x=13/4