\(\sqrt{x^2+1}+\sqrt{3\cdot x^2+16}=5-12\cdot x^2\)
TÌM X
CÁM ƠN ĐÃ GIÚP MK !
Những câu hỏi liên quan
\(\sqrt{2\cdot x^2+4\cdot x+6}\) +\(\sqrt{3\cdot x^2+6\cdot x+12}\)=5-\(2\cdot x\)-\(x^2\)
giải hệ phương trình :
a) \(\hept{\begin{cases}x\cdot\left(1+y-x\right)=-2\cdot y^2-y\\x\cdot\left(\sqrt{2\cdot y}-2\right)=y\cdot\left(\sqrt{x-1}-2\right)\end{cases}}\)
b) \(\hept{\begin{cases}1+x\cdot y+\sqrt{x\cdot y}=x\\\frac{1}{x\cdot\sqrt{x}}+y\cdot\sqrt{y}=\frac{1}{\sqrt{x}}+3\cdot\sqrt{y}\end{cases}}\)
Làm hộ mk nhé mk tick cho :))))))))))
Giải các phương trình sau
a) -x^2+4cdot x+12cdotsqrt{2cdot x+1}
b) x+sqrt{x+dfrac{1}{2}+sqrt{x+dfrac{1}{4}}}2
c) 5cdot x^2-2cdot x+1left(4cdot x-1right)cdotsqrt{x^2+1}
d) left(2cdot x-1right)cdotsqrt{10-4cdot x^2}5-2cdot x
e) sqrt{2cdot x-1}-sqrt{x+1}2cdot x-4
f) sqrt{x^2-2cdot x}+sqrt{2cdot x^2+4cdot x}2cdot x
Đọc tiếp
Giải các phương trình sau
a) \(-x^2+4\cdot x+1=2\cdot\sqrt{2\cdot x+1}\)
b) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
c) \(5\cdot x^2-2\cdot x+1=\left(4\cdot x-1\right)\cdot\sqrt{x^2+1}\)
d) \(\left(2\cdot x-1\right)\cdot\sqrt{10-4\cdot x^2}=5-2\cdot x\)
e) \(\sqrt{2\cdot x-1}-\sqrt{x+1}=2\cdot x-4\)
f) \(\sqrt{x^2-2\cdot x}+\sqrt{2\cdot x^2+4\cdot x}=2\cdot x\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
Đúng 0
Bình luận (3)
Giải phương trình \(\sqrt{x-2+\sqrt{2\cdot x+5}}+\sqrt{x+2+3\cdot\sqrt{2\cdot x-5}}=7\cdot\sqrt{2}\)
GPT : x = \(\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
giải pt
\(x=\sqrt{2-x}\cdot\sqrt{3-x}+\sqrt{3-x}\cdot\sqrt{5-x}+\sqrt{5-x}\cdot\sqrt{2-x}\)
giải pt \(2+\sqrt[3]{9x^2\cdot\left(x+2\right)}=2x+3\sqrt[3]{3x\cdot\left(x+2\right)^2}\)
giúp mk nha mk cần gấp
\(2+\sqrt[3]{b^2a}=\frac{3}{2}b+\sqrt[3]{a^2b}\)
Đúng 0
Bình luận (0)
\(\sqrt{x+3}+\sqrt{2\cdot x+4}=12-\sqrt{3\cdot x+7}\)
Đk:\(x\ge-2\)
\(pt\Leftrightarrow\sqrt{x+3}+\sqrt{2x+4}-12+\sqrt{3x+7}=0\)
\(\Leftrightarrow\sqrt{x+3}-3+\sqrt{2x+4}-4+\sqrt{3x+7}-5=0\)
\(\Leftrightarrow\frac{x+3-9}{\sqrt{x+3}+3}+\frac{2x+4-16}{\sqrt{2x+4}+4}+\frac{3x+7-25}{\sqrt{3x+7}+5}=0\)
\(\Leftrightarrow\frac{x-6}{\sqrt{x+3}+3}+\frac{2\left(x-6\right)}{\sqrt{2x+4}+4}+\frac{3\left(x-6\right)}{\sqrt{3x+7}+5}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{1}{\sqrt{x+3}+3}+\frac{2}{\sqrt{2x+4}+4}+\frac{3}{\sqrt{3x+7}+5}\right)=0\)
Dễ thấy:\(\forall x\ge2\) thì \(\frac{1}{\sqrt{x+3}+3}+\frac{2}{\sqrt{2x+4}+4}+\frac{3}{\sqrt{3x+7}+5}>0\) (loại)
Nên \(x-6=0\Rightarrow x=6\) (thỏa)
Đúng 0
Bình luận (0)
Tìm GTNN:
\(A=\sqrt{\left(x-2\right)\cdot\left(x-1\right)\cdot x\cdot\left(x+1\right)+5}\)
\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(A=\sqrt{\left(x-2\right)\left(x-1\right)x\left(x+1\right)+5}\)
\(=\sqrt{\left(x^2-x-2\right)\left(x^2-x\right)+5}\)
Đặt \(t=x^2-x\) ta đc:
\(A=\sqrt{\left(t-2\right)t+5}=\sqrt{t^2-2t+5}\)
\(=\sqrt{\left(t-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = khi \(t=1\Leftrightarrow x^2-x=1\Leftrightarrow x=\pm\frac{1}{2}+\frac{\sqrt{5}}{2}\)
Vậy....
b)\(B=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\)
\(=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(=\left|x-2\right|+\left|x+3\right|\)
Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-2\right|+\left|x+3\right|=\left|x-2\right|+\left|-x-3\right|\ge\left|x-2+\left(-x\right)-3\right|=5\)
Dấu = khi \(\left(x-2\right)\left(x+3\right)\ge0\)\(\Rightarrow-3\le x\le2\)
\(\Rightarrow\hept{\begin{cases}-3\le x\le2\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
Vậy....
Đúng 0
Bình luận (0)