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ĐK
\(\left\{{}\begin{matrix}x+3\ge0\\7-x\ge0\\2x-8\ge0\end{matrix}\right.\)
Giải hệ bất PT trên được ĐK tổng hợp là \(4\le x\le7\)
Bình phương 2 vế PT
\(x+3+7-x-2\sqrt{\left(x+3\right)\left(7-x\right)}=2x-8\)
\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(7-x\right)}=18-2x\)
BP 3 vế PT
\(4\left(x+3\right)\left(7-x\right)=324+4x^2-72x\)
\(\Leftrightarrow28x-4x^2+84-12x=324+4x^2-72x\)
\(\Leftrightarrow8x^2-88x+240=0\Leftrightarrow x^2-11x+30=0\)
Giải PT bậc 2 rồi đối chiếu với đk, bạn tự làm nốt nhé
a) đkxđ \(x\ge1\)
pt đã cho \(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+3}=0\end{matrix}\right.\)
Hiển nhiên pt thứ 2 vô nghiệm vì \(VT>0\) với mọi \(x\ge1\). Do đó pt đã cho có nghiệm duy nhất là \(x=5\)
b) đkxđ: \(x\ge-3\)
Để ý rằng \(x^2+2x+7=\left(x^2+1\right)+\left(2x+6\right)=\left(x^2+1\right)+2\left(x+3\right)\) nên nếu ta đặt \(\sqrt{x^2+1}=u\left(u\ge1\right)\) và \(\sqrt{x+3}=v\left(v\ge0\right)\) thì pt đã chot rở thành:
\(u^2+2v^2=3uv\)
\(\Leftrightarrow\left(u-v\right)\left(u-2v\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}u=v\\u=2v\end{matrix}\right.\)
Nếu \(u=v\) thì \(\sqrt{x^2+1}=\sqrt{x+3}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=x+3\end{matrix}\right.\)
Mà \(x^2+1=x+3\) \(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (nhận)
Nếu \(u=2v\) thì \(\sqrt{x^2+1}=2\sqrt{x+3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=4x+12\end{matrix}\right.\)
mà \(x^2+1=4x+12\)\(\Leftrightarrow x^2-4x-11=0\)
\(\Leftrightarrow x=2\pm\sqrt{15}\) (nhận)
Vậy pt đã cho có tập nghiệm \(S=\left\{2;-1;2\pm\sqrt{15}\right\}\)
a) \(\sqrt{2x-1}+\sqrt{x-1}=5\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{x-1}\right)^2=5^2\)
\(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow3x-2+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-1\right)}=\dfrac{27-3x}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{27-3x}{2}\ge0\\\left(2x-1\right)\left(x-1\right)=\left(\dfrac{27-3x}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}27-3x\ge0\\2x^2-2x-x+1=\dfrac{729-162x+9x^2}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\le27\\8x^2-12x+4=9x^2-162x+729\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x^2-150x+725=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x-5=0\\x-145=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=145\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=5\)
ĐK:\(x\ge\dfrac{5}{2}\)
Ta có:\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=7.2\)
\(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+6}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)
\(\Leftrightarrow2\sqrt{2x-5}=10\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow2x-5=25\Leftrightarrow2x=30\Leftrightarrow x=15\left(tm\right)\)
ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)
\(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+3}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow2.\sqrt{2x-5}+4=14\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow x=15\)
\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)
Lời giải:
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)
b.
ĐKXĐ: $x\geq \frac{3}{2}$
PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)
\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)
\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)
\(\Leftrightarrow \sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
\(đk:4\le x\le7\)
\(\sqrt{x+3}-\sqrt{7-x}=\sqrt{2x-8}\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{7-x}-\sqrt{2x-8}=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x+3}-3\right)\left(\sqrt{x+3}+3\right)}{\sqrt{x+3}-3}-\frac{\left(\sqrt{7-x}-1\right)\left(\sqrt{7-x}+1\right)}{\sqrt{7-x}+1}-\frac{\left(\sqrt{2x-8}-2\right)\left(\sqrt{2x-8}+2\right)}{\sqrt{2x-8}+2}=0\)
\(\Leftrightarrow\frac{x+3-9}{\sqrt{x+3}-3}-\frac{7-x-1}{\sqrt{7-x}+1}-\frac{2x-8-4}{\sqrt{2x-8}+2}=0\)
\(\Leftrightarrow\frac{x-6}{\sqrt{x+3}-3}+\frac{x-6}{\sqrt{7-x}+1}-\frac{2x-12}{\sqrt{2x-8}+2}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{1}{\sqrt{x+3}-3}+\frac{1}{\sqrt{7-x}+1}-\frac{2}{\sqrt{2x-8}+2}\right)=0\)
cái ngoặc thứ 2 chưa biết lập luận như nào :v
\(\Leftrightarrow x=6\left(tm\right)\)