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Có: \(a+b+c+2\sqrt{abc}=1\Rightarrow\hept{\begin{cases}a+2\sqrt{abc}=1-b-c\\b+2\sqrt{abc}=1-a-c\\c+2\sqrt{abc}=1-a-b\end{cases}}\)
\(A=\sqrt{a\left(1-b\right)\left(1-c\right)}+\sqrt{b\left(1-c\right)\left(1-a\right)}+\sqrt{c\left(1-a\right)\left(1-b\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(1-b-c+bc\right)}+\sqrt{b\left(1-a-c+ac\right)}+\sqrt{c\left(1-a-b+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}+\sqrt{b\left(b+2\sqrt{abc}+ac\right)}+\sqrt{c\left(c+2\sqrt{abc}+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a^2+2a\sqrt{abc}+abc\right)}+\sqrt{\left(b^2+2b\sqrt{abc}+abc\right)}+\sqrt{\left(c^2+2c\sqrt{abc}+abc\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a+\sqrt{abc}\right)^2}+\sqrt{\left(b+\sqrt{abc}\right)^2}+\sqrt{\left(c+\sqrt{abc}\right)^2}-\sqrt{abc}+2015\)
\(A=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2015\)
\(A=a+b+c+2\sqrt{abc}+2015\)
\(A=1+2015=2016\)
Vậy:....
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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 \(x=-\frac{3}{\sqrt{2}}\)
Lời giải:
Từ ĐKĐB suy ra tồn tại $x,y,z>0$ sao cho:
\((a,b,c)=\left(\frac{x^2}{(x+y)(x+z)}; \frac{y^2}{(y+z)(y+x)}; \frac{z^2}{(z+x)(z+y)}\right)\)
(lưu ý cách đặt ẩn phụ như thế này rất hữu ích trong các bài BĐT có điều kiện như trên)
Khi đó:
\(a(1-b)(1-c)=\frac{x^2}{(x+y)(x+z)}\left(1-\frac{y^2}{(y+z)(y+x)}\right)\left(1-\frac{z^2}{(z+x)(z+y)}\right)\)
\(=\frac{x^2}{(x+y)(x+z)}.\frac{xy+yz+xz}{(y+z)(y+x)}.\frac{xy+yz+xz}{(z+x)(z+y)}=\left(\frac{x(xy+yz+xz)}{(x+y)(y+z)(z+x)}\right)^2\)
\(\Rightarrow \sqrt{a(1-b)(1-c)}=\frac{x(xy+yz+xz)}{(x+y)(y+z)(z+x)}\)
Tương tự như vậy với các phân thức tương ứng còn lại.
\(\sqrt{abc}=\sqrt{\frac{x^2.y^2z^2}{((x+y)(y+z)(z+x))^2}}=\frac{xyz}{(x+y)(y+z)(z+x)}\)
Do đó:
\(A=\frac{x(xy+yz+xz)}{(x+y)(y+z)(z+x)}+\frac{y(xy+yz+xz)}{(x+y)(y+z)(z+x)}+\frac{z(xy+yz+xz)}{(x+y)(y+z)(z+x)}-\frac{xyz}{(x+y)(y+z)(z+x)}+2017\)
\(A=\frac{(x+y+z)(xy+yz+xz)-xyz}{(x+y)(y+z)(z+x)}+2017=\frac{(x+y)(y+z)(z+x)}{(x+y)(y+z)(z+x)}+2017=1+2017=2018\)
Cho e hỏi tí ạ ! Ta có thể dùng cách đặt ẩn này cho những trường hợp nào nữa ? Căn cứ để đặt ẩn ạ !