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a; A = (7\(x\) + 5)2 + (3\(x-5\))2 - (10 - 6\(x\)).(5 + 7\(x\))
A = 49\(x^2\) + 70\(x\) + 25 + 9\(x^2\) - 30\(x\) + 25 - 50 - 70\(x\) + 30\(x\) + 42\(x^2\)
A = (49\(x^2\) + 9\(x^2\) + 42\(x^2\)) + (70\(x-70x\)) - (30\(x\) - 30\(x\)) + (25+25-50)
A = 100\(x^2\) + 0 + 0 + (50 - 50)
A = 100\(x^2\) + 0 + 0 + 0
A = 100\(x^2\)
Thay \(x=-2\) vào A = 100\(x^2\) ta có:
A = 100.(-2)2
A = 100.4
A = 400.
2: \(N=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{39}{2}+1-3-2=\dfrac{39}{2}-4=\dfrac{31}{2}\)
3: \(P=4x^2-25-4x^2-4x-1=-4x-26\)
=-8020-26=-8046
4: \(Q=\left(y^2-9\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=y^4-81-y^4+4=-77\)
a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(A=9x\)
Thay x = 15 vào, ta có:
\(A=9.15=135\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(B=5x^2-20xy-4y^2+20xy\)
\(B=5x^2-4y\)
Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có:
\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)
c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)
\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(C=9x^2y^2-xy^3-8x^3\)
Thay \(x=\frac{1}{2};y=2\) vào, ta có:
\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)
d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)
\(D=18x^2+12x-7\)
Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
+) Với x = -2
\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
+) Với x = 2
\(D=18.2^2+12.2-7=89\)
Bài 1:
\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)
Bài 2:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
a) \(A=3x\left(x^2-2x+3\right)-x^2.\left(3x-2\right)+5\left(x^2-x\right)\)
\(=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x\)
\(=x^2+4x\)
Thay \(x=5\)vào biểu thức ta có: \(A=5^2+4.5=25+20=45\)
b) \(B=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10\); \(y=-1\)vào biểu thức ta có:
\(B=10^3-\left(-1\right)^3=1000+1=1001\)
a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
a: Đặt \(A=\left(2x+y\right)^2-2\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)^2=\left(2y\right)^2=4y^2\)
Khi y=3 thì \(A=4\cdot3^2=4\cdot9=36\)
b: Đặt \(B=\left(2x-5\right)\left(2x+5\right)-\left(2x+1\right)^2\)
\(=\left(2x\right)^2-5^2-4x^2-4x-1\)
\(=4x^2-25-4x^2-4x-1=-4x-26\)
Khi x=0 thì \(B=-4\cdot0-26=-26\)