Câu hỏi của hoàng lan ngọc

Question

1. rút gọn (1/2a+b) mũ 3+ (1/2a-b) mũ 32. tìm x biết: x mũ 3 - 3x mũ 2 +3x-1=03. cho a= ( 4x-1) mũ 3 - ( 4x-3) ( 16x mũ 2+3). Chứng minh A không phụ thuộc vào x                         ---------------...

KAl(SO4)2·12H2O · Accepted Answer

1) $\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3$$=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2$$=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3$$=\frac{a^3}{4}+3ab^2$2) $x^3-3x^2+3x-1=0$$\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0$$\Leftrightarrow\left(x+1\right)^3=0$$\Leftrightarrow x+1=0$$\Leftrightarrow x=0-1$$\Rightarrow x=-1$3) $A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)$$A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9$$A=8$Vậy: biểu thức không phụ thuộc vào biếnCâu trả lời: 1) $\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3$$=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2$$=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]$$=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3$$=\frac{a^3}{4}+3ab^2$2) $x^3-3x^2+3x-1=0$$\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0$$\Leftrightarrow\left(x+1\right)^3=0$$\Leftrightarrow x+1=0$$\Leftrightarrow x=0-1$$\Rightarrow x=-1$3) $A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)$$A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9$$A=8$Vậy: biểu thức không phụ thuộc vào biến

KAl(SO4)2·12H2O · Answer

1) $\left(x+5\right)^3-x^3-125$$=\left(x+5\right)\left(x^2+2x.5+5^2\right)-x^3-125$$=x\left(x^2+2x.5+5^2\right)+5\left(x^2+2x.5+5^2\right)-x^3-125$$=x^3+10x^2+25x+5x^2+50x+125-x^3-125$$=15x^2+75x$2) $\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0$$\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0$$\Leftrightarrow24x+10=0$$\Leftrightarrow24x=0-10$$\Leftrightarrow24x=-10$$\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}$$\Rightarrow x=-\frac{5}{12}$3) $\left(x-1\right)^3-x^3+3x^2-3x+1$$=\left(x-1\right)\left(x^2-2x+1\right)-x^3+3x^2-3x+1$$=x\left(x^2-2x+1\right)-\left(x^2-2x+1\right)-x^3+3x^2-3x+1$$=x^3-2x^2+x-x^2+2x-1-x^3-3x^2-3x+1$$=0$Vậy: biểu thức không phụ thuộc vào biếnCâu trả lời: 1) $\left(x+5\right)^3-x^3-125$$=\left(x+5\right)\left(x^2+2x.5+5^2\right)-x^3-125$$=x\left(x^2+2x.5+5^2\right)+5\left(x^2+2x.5+5^2\right)-x^3-125$$=x^3+10x^2+25x+5x^2+50x+125-x^3-125$$=15x^2+75x$2) $\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0$$\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0$$\Leftrightarrow24x+10=0$$\Leftrightarrow24x=0-10$$\Leftrightarrow24x=-10$$\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}$$\Rightarrow x=-\frac{5}{12}$3) $\left(x-1\right)^3-x^3+3x^2-3x+1$$=\left(x-1\right)\left(x^2-2x+1\right)-x^3+3x^2-3x+1$$=x\left(x^2-2x+1\right)-\left(x^2-2x+1\right)-x^3+3x^2-3x+1$$=x^3-2x^2+x-x^2+2x-1-x^3-3x^2-3x+1$$=0$Vậy: biểu thức không phụ thuộc vào biến

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