Giari PT:
\(\sqrt{x^2+2}+\sqrt{1+\frac{1}{x^2}}=\sqrt{x+3}\)
Bài 1:
Giari PT: \(\frac{1}{2}log_{\sqrt{2}}\left(x+3\right)+\frac{1}{4}log_4\left(x-1\right)^8=3log_8\left(4x\right)\)
Bài 2:
Tìm m để PT sau có nghiệm: \(x\in\left[0;1+\sqrt{3}\right]\):
\(m\left(\sqrt{x^2-2x+2}+1\right)+x\left(2-x\right)\le0\)(2)
Bài 3:
Giari HPT: \(\hept{\begin{cases}x^4-4x^2+y^2-6y+9=0\\x^2y+x^2+2y-22=0\end{cases}}\)(2)
P/s: Mình không cần gấp,cuối tuần mình mới nộp. Cac bạn gắng giúp mình nha!
1) tính
a) (\(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)với x>0 , x khác 1 b) (\(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(x-\frac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)c) \(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)với a,b>0 a #b2) Giari phương trình
\(\sqrt[3]{x-8}+\sqrt{x+7}+x^3-8x^2-8x-14=0\)
Mong mọi người giúp mình giải bài với ạ
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
Giải pt: \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}=1\)
Giải pt \(\frac{1}{\sqrt{x-1}+\sqrt{x-2}}+\frac{1}{\sqrt{x-2}+\sqrt{x-3}}+...+\frac{1}{\sqrt{x-9}+\sqrt{x-10}}=1\)
Giải pt: \(x+\sqrt[3]{x^3-x^2}+\sqrt[3]{x^3-x}=\sqrt[3]{x^2+x+\frac{1}{3}}+\sqrt[3]{x^2+\frac{1}{3}}+\sqrt[3]{x+\frac{1}{3}}\)
Giải pt
\(\sqrt{2x+\frac{2013-1}{\sqrt{2-x^2}}}-\sqrt[3]{2014-\frac{2013-1}{\sqrt{2-x^2}}}=\sqrt{x+2013}-\sqrt[3]{x+1}\)
Giải pt \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}=1\)
GIÚP MK ĐI!!!!!!!!
olm còn lỗi nên ko trình bày bth đc, bn tự viết lại nhá :))
\(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}=\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+3}+\sqrt{x+2}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}\)
\(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}=\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(\sqrt{x+2}+\sqrt{x+1}\right)\left(\sqrt{x+2}-\sqrt{x+1}\right)}\)
\(\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\)
\(VT=\sqrt{x+3}-\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+1}-\sqrt{x}\)
\(VT=\sqrt{x+3}-\sqrt{x}=1\)
Dễ r -,-
giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
giải pt \(\sqrt{2x+\frac{2013x-1}{\sqrt{2-x^2}}}-\sqrt[3]{2014-\frac{2013x-1}{\sqrt{2-x^2}}}=\sqrt{x+2003}-\sqrt[3]{x+1}\)