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6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
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\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)
\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)
Pt trở thành:
\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)
\(\Rightarrow\sqrt{1+x}+\sqrt{1-x}=2\)
\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)
\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)
b.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)
Pt trở thành:
\(t=t^2-4-16\Leftrightarrow...\)
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2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
Đk : x >= -2
pt <=> 3.\(\sqrt{\left(x+2\right).\left(x^2-2x+4\right)}\) = 2.[(x^2-2x+4)-(x+2)]
Đặt \(\sqrt{x+2}\)= a ; \(\sqrt{x^2-2x+4}\)= b ( a,b >= 0 )
pt <=> 3ab = 2a^2-2b^2
<=> 2a^2-2b^2-3ab = 0
<=> (2a^2-4ab)+(ab-2b^2) = 0
<=> (a-2b).(2a+b) = 0
<=> a=2b và 2a=-b
Từ đó bạn tự thay a,b bởi x để tìm x nha
đặt 3√x3+82(x2−3x+2)=3x3+8 = 2(x^2 - 3x +2) là (1)
DK:x≥−2x≥−2
(1)<=>2(x2−3x+2)=3√x+2.√x2−2x+4(1)<=>2(x2−3x+2)=3x+2.x2−2x+4
Đặt: √x+2=ax+2=a (a≥0)(a≥0)
√x2−2x+4=bx2−2x+4=b (b≥0)(b≥0)
=>x2−3x+2=b2−a2x2−3x+2=b2−a2
Ta có pt: 2(b2−a2)=3ab2(b2−a2)=3ab <=>(2a−b)(a+2b)=0<=>2a−b=0<=>(2a−b)(a+2b)=0<=>2a−b=0 hoặc a+2b=0a+2b=0
Đến đây bạn tự giả nhé!!!!vvv