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a, Với 7x - 1 \(\ge\)0 <=> x \(\ge\)\(\frac{1}{7}\) thì |7x - 1| = 7x - 1
lúc đó A = 2x2 + (7x - 1) - 5 + x - 2x2 = 8x - 6
Với 7x - 1 < 0 <=> x < \(\frac{1}{7}\)thì |7x - 1| = 1 - 7x
lúc đó A = 2x2 + (1 - 7x) - 5 + x - 2x2 = -6x - 4
b, Xét 2 trường hợp
TH1: x \(\ge\)\(\frac{1}{7}\) thì 8x - 6 = 2 <=> 8x = 8 <=> x = 1 (thỏa mãn)
TH2: x < \(\frac{1}{7}\) thì -6x - 4 = 2 <=> -6x = 6 <=> x = -1 (thỏa mãn)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-5\right)\left(x+8\right)-\left(x+4\right)\left(x-1\right)\)
\(=\left(x^2+3x-40\right)-\left(x^2+3x-4\right)\)
\(=x^2+3x-40-x^2-3x+4\)
\(=-36\)
b)\(x^4\left(x^2-1\right)\left(x^2+1\right)\)
\(=x^4\left(x^4-1\right)\)
\(=x^8-x^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)
Thay \(a=\frac{1}{2};x=-3\), ta có :
\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)
b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)
\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)
Thay \(a=-3;b=-4;c=2;d=3\), ta có :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
\(A=\left|2x-1\right|-\left(x-5\right)\)
\(=2x-1-x+5\)
\(=x+4\)
Để \(A=4\)
\(\Leftrightarrow x+4=4\)
\(\Rightarrow x=0\)
\(A=|2x-1|-\left(x-5\right)\)
=2x-1-x+5
=x+4
Để a = 4
=>x+4=4
<=>x=4-4=0
Vậy....