thu gọn
a)(2x-7)^2+(2x+7)^2-2(2x+7)(2x-7)
b)(5x-3)^2-(5x+3)^2-15(2x-1)
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a, \(\left(2x-7\right)^2+\left(2x+7\right)^2-2\left(2x+7\right)\left(2x-7\right)=\left(2x-7-2x-7\right)^2=\left(-14\right)^2=196\)
b, \(\left(5x-3\right)^2-\left(5x+3\right)^2-15\left(2x-1\right)\)
\(=\left(5x-3-5x-3\right)\left(5x-3+5x+3\right)-15\left(2x-1\right)\)
\(=-6.10x-15\left(2x-1\right)\)
\(=-60x-15\left(2x-1\right)=-15\left(4x+2x-1\right)=-15\left(6x-1\right)=-90x+15\)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
A=-(3x+7)+(5x-2)+(2x-10)
=-3x-7+5x-2+2x-10
=(-3x+5x+2x)-(7+2+10)
=4x-19
B = (6x+8)-(4x-5)-3x
= 6x+8-4x+5-3x
= (6x-4x-3x) + (8+5)
= -x + 13
= 13-x
C = 2(5x+3) - (2x-1) + 12
= 10x+6 - 2x + 1 + 12
= (10x-2x) + (6+1+12)
= 8x + 19
D = (x+7)-3(x+1)+2x-5
= x+7-3x-3+2x-5
= (x-3x+2x) + (7-3-5)
= -1
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
a: =>-x+2x=3-7
=>x=-4
b: =>6x+2+2x-5=0
=>8x-3=0
hay x=3/8
c: =>5x+2x-2-4x-7=0
=>3x-9=0
hay x=3
d: =>10x2-10x2-15x=15
=>-15x=15
hay x=-1
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)
\(=x^2-12x^2+16x+7x-5\)
\(=-11x^2+23x-5\)
b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)
\(=7x^3-35x-3x^3y^2+18x^2y^3\)
c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)
\(=10x^2-35x+8x-28\)
\(=10x^2-27x-28\)