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ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
`a,3x^2+7x+2=0`
`<=>3x^2+6x+x+2=0`
`<=>3x(x+2)+x+2=0`
`<=>(x+2)(3x+1)=0`
`<=>x=-2\or\x=-1/3`
d) Ta có: (x-1)(x+2)=70
\(\Leftrightarrow x^2+2x-x-2-70=0\)
\(\Leftrightarrow x^2+x-72=0\)
\(\Leftrightarrow x^2+9x-8x-72=0\)
\(\Leftrightarrow x\left(x+9\right)-8\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)
Vậy: S={8;-9}
ĐK: \(4\le x\le6\)
\(VP=\left(x-5\right)^2+2\ge2\text{ }\forall4\le x\le6\)
Đẳng thức xảy ra khi x = 5.
Với 2 số thực a, b ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
Đẳng thức xảy ra khi a = b.
\(VT^2=\left(\sqrt{x-4}+\sqrt{6-x}\right)^2\le2\left(x-4+6-x\right)=4\)
\(\Rightarrow VT\le2.\)
Đẳng thức xảy ra khi \(\sqrt{x-4}=\sqrt{6-x}\Leftrightarrow x=5.\)
Vậy ta có: \(VT\le2\le VP\)
Nên pt đã cho tương đương \(VT=2;\text{ }VP=2\Leftrightarrow x=5.\)
KL: x = 5.
<=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2=5\)
mà \(\sqrt{3\left(x+1\right)^2+9}\ge3\), \(\sqrt{5\left(x^2-1\right)^2+4}\ge4\), \(2\left(x+1\right)^2\ge0\)với mọi x
=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2\ge3+2+0=5\)
'=" xảy ra<=> x+1=0<=> x=-1
Ta có: \(x^5-x^4+3x^3+3x^2-x+1=0\)
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
a ) Đặt \(\sqrt{x+1}=a\Rightarrow x+1=a^2\Rightarrow x=a^2-1\)
Ta có : \(x^2+x+12\sqrt{x+1}=36\)
\(\Leftrightarrow x\left(x+1\right)+12a=36\)
\(\Leftrightarrow a^2\left(a^2-1\right)+12a-36=0\)
\(\Leftrightarrow a^4-a^2+12a-36=0\)
\(\Leftrightarrow a^3\left(a-2\right)+2a^2\left(a-2\right)+3a\left(a-2\right)+18\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2+3a+18\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left[a^2\left(a+3\right)-a\left(a+3\right)+6\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\left(a^2-a+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+1}=-3\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x+1=4\Leftrightarrow x=3\)
Vậy ...
b ) \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow\left(x^4-8x^2+16\right)+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(4-x^2=a\) , ta có :
\(a^2+x-4=0\) \(\Rightarrow x=4-a^2\)
Ta có : x = \(4-a^2;a=4-x^2\)
\(\Leftrightarrow x-a=x^2-a^2\)
\(\Leftrightarrow\left(x-a\right)\left(1-x-a\right)=0\)
\(\Leftrightarrow\left(x-4+x^2\right)\left(1-x-4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2+x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow...\)