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\(ĐK:x\ge-3\\ PT\Leftrightarrow\sqrt{x-3}=2\Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
Bài 1:
a) \(A\left(x\right)+B\left(x\right)=\left(-x^3+x^2-5x+1\right)+\left(x^3+4x-5\right)\)
\(=-x^3+x^2-5x+1+x^3+4x-5\)
\(=\left(-x^3+x^3\right)+x^2+\left(-5x+4x\right)+\left(1-5\right)\)
\(=x^2-x-4\)
b) \(A\left(x\right)-B\left(x\right)=\left(-x^3+x^2-5x+1\right)-\left(x^3+4x-5\right)\)
\(=-x^3+x^2-5x+1-x^3-4x+5\)
\(=\left(-x^3-x^3\right)+x^2+\left(-5x-4x\right)+\left(1+5\right)\)
\(=-2x^3+x^2-9x+6\)
Bài 2
* \(P+Q=\left(x^5+7x^3+1\right)+\left(x^3-4x^5+2\right)\)
\(=x^5+7x^3+1+x^3-4x^5+2\)
\(=\left(x^5-4x^5\right)+\left(7x^3+x^3\right)+\left(1+2\right)\)
\(=-3x^5+8x^3+3\)
* \(P-Q=\left(x^5+7x^3+1\right)-\left(x^3-4x^5+2\right)\)
\(=x^5+7x^3+1-x^3+4x^5-2\)
\(=\left(x^5+4x^5\right)+\left(7x^3-x^3\right)+\left(1-2\right)\)
\(=5x^5+6x^3-1\)
Ta có: lx-1l + l4-xl = 3 <=> lx-1l + lx-4l = 3
TH1: Nếu x < 1, ta có: TH2: Nếu 1 < x < 4, ta có: TH3: Nếu x > 4, ta có: 1 - x + 4 - x = 3 x - 1 + 4 - x = 3 x - 1 + x - 4 = 3 <=>5 - 2x = 3 <=> 3 =3 (TM) <=> 2x - 5 = 3
<=> 2x = 5 - 3 = 2 <=> x = 1;2;3;4 <=> 2x = 3 + 5 = 8 <=> x = 1 (TM) < => x = 4(TM) Vậy x = 1;2;3;4.
Đặt : \(P=\frac{48^2\cdot8^5\cdot100^9}{12^2\cdot2^{15}\cdot4^2}\)
\(=\frac{\left(2^4\cdot3\right)^2\cdot\left(2^3\right)^5\cdot\left(2^2\cdot5^2\right)^9}{\left(2^2\cdot3\right)^2\cdot2^{15}\cdot\left(2^2\right)^2}\)
\(=\frac{2^8\cdot3^2\cdot2^{15}\cdot2^{18}\cdot5^{18}}{2^4\cdot3^2\cdot2^{15}\cdot2^4}\)
\(=\frac{2^{41}\cdot3^2\cdot5^{18}}{2^{23}\cdot3^2}=2^{18}\cdot5^{18}=\left(2\cdot5\right)^{18}=10^{18}\)
Vậy : \(P=10^{18}\)
đầu bài là như này đúng không hả bạn
\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
Ta có :\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
\(\frac{2}{3}:\left(x-1\right)\)\(=\frac{1}{4}\)
\(\left(x-1\right)\)\(=\frac{8}{3}\)
\(x=\frac{11}{3}\)
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