Phân tích ra thừa số:
a, x - 9 với x > 0
b, x - 5\(\sqrt{x}\) + 4
d, x-2\(\sqrt{x-1}\) - a2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)
c) x - 6\(\sqrt{x}\)+ 9 = \(\left(\sqrt{x}\right)^2\)- 2.\(\sqrt{x}\).3 + 9 = \(\left(\sqrt{x}-3\right)^2\)
d) Tương tự.
a,b) Không hiểu
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x-6\sqrt{x}+9=\left(\sqrt{x}-3\right)^2\)
\(d,x-4\sqrt{x}+4=\left(\sqrt{x}-2\right)^2\)
b4 :
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)
b5:
\(a,ĐK:x\ge1\)
\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)
\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:x\ge5\)
\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)
\(\Leftrightarrow-5\sqrt{x-5}=2\)
\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)
\(c,ĐK:x>0\)
\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)
\(\Leftrightarrow x+9=6\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
a: \(2018-x=\left(\sqrt{2018}-\sqrt{x}\right)\left(\sqrt{2018}+\sqrt{x}\right)\)
b: \(2018+x=2018-\left(-x\right)=\left(\sqrt{2018}-\sqrt{-x}\right)\left(\sqrt{2018}+\sqrt{-x}\right)\)
c: \(7x-12\sqrt{x}+5\)
\(=7x-7\sqrt{x}-5\sqrt{x}+5\)
\(=\left(\sqrt{x}-1\right)\left(7\sqrt{x}-5\right)\)
\(a,=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\\ b,=\left(\sqrt{x}-\sqrt{y}\right)^2\\ c,=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\\ d,=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)\\ =\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)
Bài 2:
a: Ta có: \(\sqrt{\sqrt{5}-x\sqrt{3}}=\sqrt{8+2\sqrt{15}}\)
\(\Leftrightarrow\sqrt{5}-x\sqrt{3}=8+2\sqrt{15}\)
\(\Leftrightarrow x\sqrt{3}=\sqrt{5}-8-2\sqrt{15}\)
\(\Leftrightarrow x=\dfrac{\sqrt{15}-8\sqrt{3}-6\sqrt{5}}{3}\)
b: Ta có: \(\sqrt{2+\sqrt{\sqrt{x}+3}}=3\)
\(\Leftrightarrow\sqrt{\sqrt{x}+3}=7\)
\(\Leftrightarrow\sqrt{x}=46\)
hay x=2116
Bài đã chữa:
a, x - 9 (x > 0)
= \(\left(\sqrt{x}\right)^2\) - 32
= \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\)
b, x - 5\(\sqrt{x}\) + 4
= x - \(4\sqrt{x}-\sqrt{x}\) +4
= \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)\)
Câu d sai đề kìa bạn