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a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
Bài 6
\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab\)
Bài 5 :
\(a,16x^2-\left(4x-5\right)^2=15\)
\(16x^2-16x^2+40x-25-15=0\)
\(40x-40=0\)
\(40x=40\)
\(x=1\)
\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(4x^2+12x+9-4x^2+4=49\)
\(12x=36\)
\(x=3\)
\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(4x^2-1+1-4x+4x^2=18\)
\(8x^2-4x-18=0\)
\(2\left(4x^2-2x-9\right)=0\)
\(x=\frac{1-\sqrt{37}}{4}\)
\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(12x=4\)
\(x=\frac{1}{3}\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
Bài 7
\(a,A=x^2-2x+5\)
\(=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
GTNN \(A=4\) khi \(\left(x-1\right)^2=0\Rightarrow x=1\)
\(b,B=x^2-x+1\)
\(=\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(c,C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x=t\)
\(\Rightarrow C=\left(t-6\right)\left(t+6\right)\)
\(=t^2-36\)
\(\left(x^2+5x\right)^2-36\ge36\forall x\)
\(d,D=x^2+5y^2-2xy+4y-3\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)-4\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2-4\ge-4\)
\(a,\left(x+2\right)^2=x^2+4x+4\)
\(b,\left(x-1\right)^2=x^2-2x+1\)
\(c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
\(d,\left(x^3+2y^2\right)^2=x^6+4x^3y^2+4y^4\)
1) \(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x\)
\(=2x-5x^3-x^2\)
2) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left(6x^2+23x-55\right)-\left(6x^2+23x+21\right)\)
\(=-76\)
Làm lại câu 1
\(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x^2\)
\(=x-5x^3\)
Bài làm:
\(A=\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(y-x\right)\)
\(A=\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(A=\left(x+y+x-y\right)^2\)
\(A=\left(2x\right)^2\)
Với x = -1/3 ta được:
\(A=\left(2.\frac{-1}{3}\right)^2=\frac{4}{9}\)
A=(x+y)2+(x-y)2-2(x+y)(y-x)
A=(x+y)2+2(x+y)(x-y)+(x-y)2
A=(x+y+x-y)2
A=(2x)2
với x+-1/3 ta được:
A=(2.-1/3)2=4/9
làm nhiều rồi
hehe
hihi
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
Bài 3 :
\(a,A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(=x^2-2xy+y^2+x^2+2xy+y^2\)
\(=2x^2+2y^2\)
\(b,B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(=8ab\)
\(c,C=\left(x+y\right)^2-\left(x-y\right)^2\)
Tương tự bài câu B
\(=4xy\)
\(d,D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+1-8x^2+24x-18+4\)
\(=-4x^2+20x-13\)
\(a,A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(A=2x^2+2y^2\)
\(b,B=2b^2\)
\(c,C=2y^2\)
Bài 3: a) A = (x - y)2 + (x + y)2 = x2 - 2xy + y2 + x2 + 2xy + y2 = 2x2 + 2y2
b) (2a + b)2 - (2a - b)2 = (2a + b - 2a + b)(2a + b + 2a - b) = 2b.4a = 8ab
c) (x + y)2 - (x -y)2 = (x + y - x + y)(x + y +x - y) = 2y.2x = 4xy
d) (2x - 1)2 - 2(2x - 3)2 + 4 = 4x2 - 4x + 1 - 2(4x2 - 12x + 9) + 4 = 4x2 - 4x + 1 - 8x2 + 24x - 18 + 4 = -4x2 + 20x - 12
Bài 4: a) A = (x + 3)2 + (x - 3)(x + 3) - 2(x + 2)(x - 4)
A = x2 + 6x + 9 + x2 - 9 - 2(x2 - 4x + 2x - 8)
A = 2x2 + 6x - 2x2 + 4x + 16
A = 10x + 16
=> A = 10.(-1/2) + 16 = 11
b) B = (3x + 4)2 - (x - 4)(x + 4) - 10x = 9x2 + 24x + 16 - x2 + 16 - 10x = 8x2 + 12x + 32
=> B = 8.(-1/10)2 + 12.(-1/10) + 32 = 2/25 - 6/5 + 32 = 772/25
c) C = (x + 1)2 - (2x - 1)2 + 3(x - 2)(x + 2)
C = x2 + 2x + 1 - 4x2 + 4x - 1 + 3x2 - 12
C = 6x - 12
=> C = 6.1 - 12 = 6 -12 = -6
d) D = (x - 3)(x + 3) + (x - 2)2 - 2x(x - 4)
D = x2 - 9 + x2 - 4x + 4 - 2x2 + 8x
D = 4x - 5
=> D = 4.(-1) -5 = -4 - 5 = -9
Bài 4 ;
\(a,A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right)\)
\(=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)\)
\(=2x^2+6x-2x^2+4x+16\)
\(=10x+16\)
Thay \(x=-\frac{1}{2}\)vào bt A ta được :
\(A=10\cdot\left(-\frac{1}{2}\right)+16\)
\(=-5+16=11\)
\(b,B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)
\(=9x^2+24x+16-x^2+16-10x\)
\(=8x^2-14x+32\)
\(=8\cdot\left(-\frac{1}{10}\right)-14\cdot\left(-\frac{1}{10}\right)+32=\frac{163}{5}\)
\(c,C=x^2+2x+1-4x^2+4x-1+3x^2-12\)
\(=6x-12\)
\(=6-12=-6\)
\(d,D=x^2-9+x^2-4x+4-2x^2+8\)
\(=-4x+3\)
\(=4-3=1\)
4/
a/ \(A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right).\)
\(A=x^2+6x+9+x^2-9-2\left(x^2-4x+2x-8\right)\)
\(A=x^2+6x+9+x^2-9-2x^2+8x-4x+16\)
\(A=10x+16\) thay x = -1/2
\(A=10\cdot-\frac{1}{2}+16=11\)
b/ \(\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)
\(=9x^2+24x+16-\left(x^2-16\right)-10x\)
\(=9x^2+24x+16-x^2+16-10x\)
\(=8x^2+14x+32\)
thay x= -1/10
\(=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\left(-\frac{1}{10}\right)+32\)
\(=\frac{2}{25}-\frac{7}{5}+32=\frac{767}{25}\)
c/ \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)
\(C=x^2+2x+1-\left(4x^2-2x+1\right)+3\left(x^2-4\right)\)
\(C=x^2+2x+1-4x^2+2x-1+3x^2-12\)
\(C=4x-12\) thay x=1
\(C=4\cdot1-12=-8\)
d/ \(D=\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2-2x\left(x-4\right)\)
\(D=x^2-9+x^2-2x+4-2x^2+8x\)
\(D=6x-5\) thay x= -1
\(D=6\cdot\left(-1\right)-5=-11\)
Bài 3:
a, \(A=\left(x-y\right)^2+\left(x+y\right)^2\)
\(=x^2-2xy+y^2+x^2+2xy+y^2\)
\(=2x^2+2y^2\)
\(=2\left(x^2+y^2\right)\)( Rút gọn thế này đc chưa :V )
b, \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=4a^2+2\cdot2a\cdot b+b^2-4a^2+2\cdot2a\cdot b-b^2\)
\(=\left(4a^2-4a^2\right)+\left(b^2-b^2\right)+4ab+4ab\)
\(=8ab\)
c, \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=2xy+2xy\)
\(=4xy\)
d, \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x-2\cdot\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x-8x^2+24x-18+4\)
\(=-4x^2+20x-14\)
P/s: Chả chắc đc câu nào 100% :v Rối. Bài 4 sẽ có trong ít phút nữa :)