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Đặt \(\hept{\begin{cases}\sqrt{2x^2+7x+10}=a\left(a>0\right)\\\sqrt{2x^2+x+4}=b\left(b>0\right)\end{cases}}\)
Ta có \(a^2-b^2=6x+6\)
Từ đó PT ban đầu thành
\(a+b=\frac{a^2-b^2}{2}\)
\(\Leftrightarrow2\left(a+b\right)-\left(a^2-b^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(2-a+b\right)=0\)
\(\Leftrightarrow a=2+b\)
\(\Leftrightarrow\sqrt{2x^2+7x+10}=2+\sqrt{2x^2+x+4}\)
\(\Leftrightarrow3x+1=2\sqrt{2x^2+x+4}\)
\(\Leftrightarrow x^2+2x-15=0\)
\(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Đặt \(\sqrt{2x^2+7x+10}=a;\sqrt{2x^2+x+4}=b\left(a,b>0\right)\)
pt <=> a + b = 3(x + 1)
Mà a2 - b2 = 2x2 + 7x + 10 - 2x2 - x - 4 = 6x + 6
nên pt <=> a + b = \(\dfrac{a^2-b^2}{2}\)
<=> (a - b)(a + b) = 2(a + b)
Vì a;b > 0 nên a + b khác 0. Chia cả 2 vế của pt cho a + b ta có
pt <=> a - b = 2
<=> \(\sqrt{2x^2+7x+10}-\sqrt{2x^2+x+4}=2\)
<=> \(\sqrt{2x^2+7x+10}=2+\sqrt{2x^2+x+4}\)
Bình phương 2 vế ta có:
pt <=> \(2x^2+7x+10=2x^2+x+8+8\sqrt{2x^2+x+4}\)
<=> \(3x+1=4\sqrt{2x^2+x+4}\)
Bình phương lần nữa rồi làm nốt, làm xong thì thử lại.
Câu 1:
ĐK: \(x\geq -2\)
Đặt \(\sqrt{x+5}=a; \sqrt{x+2}=b(a,b\geq 0)\)
\(\Rightarrow ab=\sqrt{(x+5)(x+2)}=\sqrt{x^2+7x+10}\)
PT trở thành:
\((a-b)(1+ab)=3\)
\(\Leftrightarrow (a-b)(1+ab)=(x+5)-(x+2)=a^2-b^2\)
\(\Leftrightarrow (a-b)(1+ab)-(a-b)(a+b)=0\)
\(\Leftrightarrow (a-b)(1+ab-a-b)=0\)
\(\Leftrightarrow (a-b)(a-1)(b-1)=0\)
Vì \(a\neq b\Rightarrow \left[\begin{matrix} a-1=0\\ b-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} a=\sqrt{x+5}=1\\ b=\sqrt{x+2}=1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-4\\ x=-1\end{matrix}\right.\). Vì $x\geq -2$ nên chỉ có $x=-1$ là nghiệm duy nhất.
Câu 2:
ĐK: \(-4\leq x\leq 4\)
Ta có: \((\sqrt{x+4}-2)(\sqrt{4-x}+2)=2x\)
\(\Leftrightarrow \frac{(x+4)-2^2}{\sqrt{x+4}+2}.(\sqrt{4-x}+2)=2x\)
\(\Leftrightarrow x.\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow x\left(\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}-2\right)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ \sqrt{4-x}+2=2\sqrt{x+4}+4(*)\end{matrix}\right.\)
Xét $(*)$
Đặt \(\sqrt{4-x}=a; \sqrt{x+4}=b\) thì ta có hệ:
\(\left\{\begin{matrix} a^2+b^2=8\\ a+2=2b+4\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^2+b^2=8\\ a=2(b+1)\end{matrix}\right.\)
\(\Rightarrow 4(b+1)^2+b^2=8\)
\(\Leftrightarrow 5b^2+8b-4=0\Leftrightarrow (5b-2)(b+2)=0\)
\(\Rightarrow b=\frac{2}{5}\) (do \(b\geq 0)\)
\(\Rightarrow x+4=b^2=\frac{4}{25}\Rightarrow x=\frac{-96}{25}\) (t/m)
Vậy \(x\in \left\{ \frac{-96}{25}; 0\right\}\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) ĐK: x>=-2
=> \(\sqrt{x+5}+\sqrt{x+2}>0\)
Nhân liên hợp:
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
<=> \(\left(x+5-x-2\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
<=> \(3\left(1+\sqrt{x^2+7x+10}\right)=3\)
<=>1+\(\sqrt{\left(x+5\right)\left(x+2\right)}=1\)
<=> \(\sqrt{\left(x+5\right)\left(x+2\right)}=0\)
<=> (x+5) (x+2) =0
<=> x=-5 hoac x=-2
-Do x>= -2.
Vay x=-2
1. \(x^4-x^2+3x+5=2\sqrt{x+1}\) ĐK: \(x\ge-1\)
\(\Leftrightarrow\left(x^4-x^2+2x+2\right)+\left(x+1-2\sqrt{x+1}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-2x+2\right)+\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\left[\left(\sqrt{x}+1\right)^2\left(x^2-2x+2\right)+1\right]=0\)
Dễ thấy \(\left(\sqrt{x}+1\right)^2\left(x^2-2x+2\right)+1>0\)
Vậy x =1
3. ĐK: \(x\ge-2\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+5}\ge0\\b=\sqrt{x+2}\ge0\end{matrix}\right.\)
pt trên được viết lại thành
\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=\sqrt{x+2}\\\sqrt{x+5}=1\\\sqrt{x+2}=1\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
Ta có: \(\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}=3x+3\)
\(\Rightarrow\sqrt{2x^2+7x+10}+\sqrt{2x^2+x+4}-3x-3=0\)
\(\Rightarrow\sqrt{2x^2+7x+10}-7+\sqrt{2x^2+x+4}-5-3x+9=0\)
\(\Rightarrow\frac{2x^2+7x+10-49}{\sqrt{2x^2+7x+10}+7}+\frac{2x^2+x+4-25}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)
\(\Rightarrow\frac{\left(x-3\right)\left(2x+13\right)}{\sqrt{2x^2+7x+10}+7}+\frac{\left(x-3\right)\left(2x+7\right)}{\sqrt{2x^2+x+4}+5}-3\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}+5}-3\right)=0\)
mà \(\frac{2x+13}{\sqrt{2x^2+7x+10}+7}+\frac{2x+7}{\sqrt{2x^2+x+4}}-3< 0\)
=> x - 3 = 0 => x = 3
Vậy x = 3
dân dương ơi bài này dễ mà nhân liên hợp là ok