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\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(=9x\)
Thay x=15 \(\Rightarrow A=9.15=135\)
\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)
\(=19x^2y^2-11xy^3-8x^3\)
Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)
\(=19-44-1\)
\(=-26\)
Câu 1 :
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:
\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)
\(\Leftrightarrow\left(x+2\right)^2=-2\)
Ta có:
\(\left(x+2\right)^2\ge0\forall x\)
Vậy pt vô nghiệm
Vậy:ko......
Câu 3:
\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)
\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)
Câu 4:
\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)
\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)
câu 6,
Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)
\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)
\(\Rightarrow-4x+3=8\)
\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)
Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)
\(\Rightarrow7x^2=-x\)
\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).
Bài làm:
a) \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)
\(=3x^2+15x-3x^2+3x-18x+18\)
\(=18\)=> không phụ thuộc GT biến
b) \(2x\left(x+3\right)-\left(x-5\right)\left(7+2x\right)\)
\(=2x^2+6x-7x-2x^2+35+10x\)
\(=9x+35\)=> có phụ thuộc GT biến
c) \(5x\left(x^2-7x+2\right)-x^2\left(5x-8\right)+27x^2-10x\)
\(=5x^3-35x^2+10x-5x^3+8x^2+27x^2-10x\)
\(=0\)=> không phụ thuộc GT biến
cho mk hỏi tại sao chỗ (3x+18)(x-1) bạn lại ra được 3x2+3x -18x+18
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
\(5x^2+5y^2+8xy+2x-2y+2=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-2y+1\right)+4\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+4\left(x+y\right)^2=0\)
\(\Rightarrow x=-1;y=1\)
Khi đó:
\(M=\left(1-1\right)^{2010}+\left(2-1\right)^{2011}+\left(1-1\right)^{2012}\)
\(=1\)
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)