Question

[tran phuong thao]

1.giải phương trình

\(\frac{9}{x^2}\)+\(\frac{2x}{\sqrt{2x^2+9}}\)=1

2, cho a,b,c >0 thỏa mãn

a+b+c+\(\sqrt{a.b.c}\)=4

tính giá trị P=\(\sqrt{a.\left(4-b\right).\left(4-c\right)}\) +\(\sqrt{b.\left(4-c\right).\left(4-a\right)}\) +\(\sqrt{c.\left(4-a\right).\left(4-b\right)}\) -\(\sqrt{abc}\)

các bạn giúp mình nhé mình cảm ơn

Answer 1

[Trần Việt Linh]

1) \(\frac{9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}=1\left(ĐK:x\ne0\right)\)

Đặt: \(\sqrt{2x^2+9}=a\left(a\ge0\right)\)

\(\Leftrightarrow2x^2+9=a^2\Leftrightarrow9=a^2-2a^2\)

Khi đó pt đã cgo trở rhanhf:

\(\frac{a^2-2x^2}{x^2}+\frac{2x}{a}=1\)

\(\Leftrightarrow\left(\frac{a}{x}\right)^2-2+\frac{2x}{a}-1=0\)

\(\Leftrightarrow\left(\frac{a}{x}\right)^2+\frac{2x}{a}-3=0\) (*)

Đặt: \(\frac{a}{x}=b\) khi đó (*) trở thành:

\(b^2+\frac{2}{b}-3=0\)

\(\Leftrightarrow b^3+2-3b=0\)

\(\Leftrightarrow\left(b^3-b\right)-\left(2b-2\right)=0\)

\(\Leftrightarrow b\left(b-1\right)\left(b+1\right)-2\left(b-1\right)=0\)

\(\Leftrightarrow\left(b-1\right)\left(b^2+b-2\right)=0\)

\(\Leftrightarrow\left(b-1\right)^2\left(b+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}b-1=0\\b+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}b=1\\b=-2\end{array}\right.\)

Với: \(b=1\) ta có:

\(\frac{a}{x}=1\Leftrightarrow a=x\Leftrightarrow\sqrt{2x^2+9}=x\Leftrightarrow2x^2+9=x^2\Leftrightarrow x^2+9=0\left(loai\right)\)

Với: \(b=-2\) ta có:

\(\frac{a}{x}=-2\)

\(\Leftrightarrow a=-2x\)

\(\Leftrightarrow\sqrt{2x^2+9}=-2x\)

\(\Leftrightarrow2x^2+9=4x^2\)

\(\Leftrightarrow2x^2=9\)

\(\Leftrightarrow x^2=\frac{9}{2}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{\sqrt{2}}\\x=-\frac{3}{\sqrt{2}}\end{array}\right.\)

Thử lại ta thấy: \(x=\frac{3}{\sqrt{2}}\left(ktm\right);x=-\frac{3}{\sqrt{x}}\left(tm\right)\)

Vaayk pt đã cho có nhgieemj là \(x=-\frac{3}{\sqrt{2}}\)

Answer 2

[tran phuong thao]

cảm ơn bạn nhìu