phân tích đa tức thành nhân tử
a) 5+ \(\sqrt{x}\) + 25 - x
b) xy -x\(\sqrt{y}\) + \(\sqrt{y}\) - 1
c)\(\sqrt{a-b}\) - \(\sqrt{a^2-b^2}\)
d) \(\sqrt{ax}\) + \(\sqrt{by}\) - \(\sqrt{bx}\) -\(\sqrt{ay}\)
Giair hộ mình vs ạ!
phân tích đa thức thành nhân tử (với a b x y không âm, a> b)
a) xy - \(y\sqrt{x}\) + \(\sqrt{x}-1\)
b) \(\sqrt{ab}-\sqrt{by}+\sqrt{bx}+\sqrt{ay}\)
c) \(\sqrt{a+b}+\sqrt{a^2+b^2}\)
d) 12 - \(\sqrt{x}\) - x
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
Phân tích thành nhân tử ( với các số x, y, a, b không âm và \(a\ge b\))
a) \(xy-y\sqrt{x}+\sqrt{x}-1\)
b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
c) \(\sqrt{a+b}+\sqrt{a^2-b^2}\)
d) \(12-\sqrt{x}-x\)
\(a)\) \(xy-y\sqrt{x}+\sqrt{x}-1\)
= \(y\sqrt{x}.(\sqrt{x}-1)+\sqrt{x}-1\)
=\((\sqrt{x}-1).(y\sqrt{x}+1)\).
\(b)\)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
=\(\sqrt{a}.\sqrt{x}-\sqrt{b}.\sqrt{y}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}\)
=\(\sqrt{a}.\sqrt{x}+\sqrt{b}.\sqrt{x}-\sqrt{a}.\sqrt{y}-\sqrt{b}.\sqrt{y}\)
=\(\sqrt{x}.(\sqrt{a}+\sqrt{b})-\sqrt{y}.(\sqrt{a}+\sqrt{b})\)
=\((\sqrt{x}-\sqrt{y}).(\sqrt{a}+\sqrt{b})\).
\(c)\)\(\sqrt{a+b}+\sqrt{a^2-b^2}\)
=\(\sqrt{a+b}+\sqrt{(a+b).(a-b)}\)
=\(\sqrt{a+b}+\sqrt{a+b}.\sqrt{a-b}\)
=\(\sqrt{a+b}.\left(1+\sqrt{a-b}\right)\).
\(d)\) \(12-\sqrt{x}-x\)
=\(12-4\sqrt{x}+3\sqrt{x}-x\)
=\(4.\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
=\(\left(3-\sqrt{x}\right).\left(4+\sqrt{3}\right)\).
a) \(xy-y\sqrt{x}+\sqrt{x}-1=\left(\sqrt{x}\right)^2.y-y\sqrt{x}+\sqrt{x}-1\)
\(=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)
b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}=\left(\sqrt{ax}+\sqrt{bx}\right)-\left(\sqrt{ay}+\sqrt{by}\right)\)
\(=\sqrt{x}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{y}\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
c) \(\sqrt{a+b}+\sqrt{a^2-b^2}=\sqrt{a+b}+\sqrt{\left(a-b\right)\left(a+b\right)}\)
\(=\sqrt{a+b}+\sqrt{a-b}.\sqrt{a+b}\)
\(=\sqrt{a+b}.\left(1+\sqrt{a-b}\right)\)
d) \(12-\sqrt{x}-x=12-\sqrt{4x}+\sqrt{3x}-x\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(3-\sqrt{x}\right)\left(4+\sqrt{x}\right)\)
Phân tích thành nhân tử:
a) \(xy-y\sqrt{x}+\sqrt{x}-1\)
b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
a> = \(y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(y\sqrt{x}-1\right)\)
a) \(xy-y\sqrt{x}+\sqrt{x}-1\)
\(=y\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(y\sqrt{x}+1\right)\)
b) \(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
\(=\left(\sqrt{ax}-\sqrt{ay}\right)+\left(-\sqrt{by}+\sqrt{bx}\right)\)
\(=\sqrt{a}.\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
Phân tích thành nhân tử
a) \(\sqrt{a^3+b^3}+\sqrt{a^2-b^2}\)
b)\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{xy}\)
c) \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
d) \(a+5\sqrt{a}+4\)
\(\text{a) }\sqrt{a^3+b^3}+\sqrt{a^2-b^2}=\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}+\sqrt{\left(a+b\right)\left(a-b\right)}\)
\(=\sqrt{a+b}\left(\sqrt{a^2-ab+b^2}+\sqrt{a-b}\right)\)
\(\text{b) }\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{xy}\text{ không phân tích được.}\)
\(\text{c) }=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\left(\sqrt{x}-\sqrt{y}\right).\sqrt{xy}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+2\sqrt{xy}\right)\)\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(\text{d) }a+5\sqrt{a}+4=\sqrt{a}.\sqrt{a}+\sqrt{a}+4\sqrt{a}+4=\sqrt{a}\left(\sqrt{a}+1\right)+4\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+4\right)\)
Tìm điều kiện xác định và phân tích các đa thức sau thành nhân tử:
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(D=\sqrt{x^2+3x+2}+\sqrt{x+1}+2\sqrt{x+2}+2\)
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
\(ĐKXĐ:x;y\ge0\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\left(\sqrt{x^3}+\sqrt{x^2y}\right)-\left(\sqrt{y^3}+\sqrt{xy^2}\right)\)
\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)\)
Phân tích đa thức thành nhân tử (với các căn thức đã cho đều có nghĩa)
A = \(x-y-3\left(\sqrt{x}+\sqrt{y}\right)\)
B = \(x-4\sqrt{x}+4\)
C = \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
D = \(5x^2-7x\sqrt{y}+2y\)
Phân tích đa thức thành nhân tử (với các căn thức đều đã có nghĩa):
a) A = \(\sqrt{x^3}\) - \(\sqrt{y^3}\) + \(\sqrt{x^2y}\) - \(\sqrt{xy^2}\)
b) B = 5x2 - 7x\(\sqrt{y}\) + 2y
a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
b: \(B=5x^2-7x\sqrt{y}+2y\)
\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)
\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)
Phân tích đa thức thành nhân tử
a,\(xy+y\sqrt{xy}+\sqrt{x}\sqrt{y}\)
b,\(6\sqrt{xy}+6xy-4x\sqrt{x}-9y\sqrt{y}\)
c,\(x+2y\sqrt{x}-3y^2\)
d,a\(a\sqrt{a}-2b\sqrt{b}-3b\sqrt{a}\)
1/ \(ab+b\sqrt{a}+\sqrt{a}+1\)
2/ \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
3/ \(xy-y\sqrt{x}+\sqrt{x}-1\)
4/\(\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)
5/ \(\sqrt{a+b}+\sqrt{a^2+b^2}\)
6/\(12-\sqrt{x}-x\)