Giai pt
\(\sqrt{1-\sqrt{x^2-x}=\sqrt{x}-1}\)
Những câu hỏi liên quan
Giai pt:
\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\)
ĐK:x\(\ge2\)\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+1}-\sqrt{x-2-2\sqrt{x}-2+1}=1\Leftrightarrow\sqrt{\left(\sqrt{x-2}+1\right)^2}-\sqrt{\left(\sqrt{x-2}-1\right)^2}=1\Leftrightarrow\left|\sqrt{x-2}+1\right|-\left|\sqrt{x-2}-1\right|=1\Leftrightarrow\sqrt{x-2}+1-\left|\sqrt{x-2}-1\right|=1\)(1)
TH1: nếu \(\sqrt{x-2}< 1\Leftrightarrow x-2< 1\Leftrightarrow x< 3\) và x>2 thì
(1)⇔\(\sqrt{x-2}+1-1+\sqrt{x-2}=1\Leftrightarrow2\sqrt{x-2}=1\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\Leftrightarrow x-2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)TH2: nếu \(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\) thì
(1)\(\Leftrightarrow\sqrt{x-2}+1-\sqrt{x-2}+1=1\Leftrightarrow2=1\left(ktm\right)\)
Vậy S={\(\dfrac{9}{4}\)}
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Giai pt:
a. \(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)
b. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
a/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)
\(\Leftrightarrow x-1=8x-3+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)
Do \(x\ge1\Rightarrow2-7x< 0\Rightarrow\left\{{}\begin{matrix}VP\ge0\\VT< 0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|=2\)
Mà \(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)
Dấu "=" xảy ra khi và chỉ khi \(1-\sqrt{x-1}\ge0\Rightarrow x\le2\Rightarrow1\le x\le2\)
Vậy nghiệm của pt là \(1\le x\le2\)
1. Cho pt: x2 -2(m+1)x+m20 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2m+2.
2. Giai pt: left(x-1right)sqrt{2left(x^2+4right)}x^2-x-2
3. Giai hệ pt: left{{}begin{matrix}frac{1}{sqrt[]{x}}-frac{sqrt{x}}{y}x^2+xy-2y^2left(1right)left(sqrt{x+3}-sqrt{y}right)left(1+sqrt{x^2+3x}right)3left(2right)end{matrix}right.
4. Giai pt trên tập số nguyên x^{2015}sqrt{yleft(y+1right)left(y+2right)left(y+3right)}+1
Đọc tiếp
1. Cho pt: x2 -2(m+1)x+m2=0 (1). Tìm m để pt có 2 nghiệm x1 ; x2 thỏa mãn (x1-m)2 + x2=m+2.
2. Giai pt: \(\left(x-1\right)\sqrt{2\left(x^2+4\right)}=x^2-x-2\)
3. Giai hệ pt: \(\left\{{}\begin{matrix}\frac{1}{\sqrt[]{x}}-\frac{\sqrt{x}}{y}=x^2+xy-2y^2\left(1\right)\\\left(\sqrt{x+3}-\sqrt{y}\right)\left(1+\sqrt{x^2+3x}\right)=3\left(2\right)\end{matrix}\right.\)
4. Giai pt trên tập số nguyên \(x^{2015}=\sqrt{y\left(y+1\right)\left(y+2\right)\left(y+3\right)}+1\)
1.Giai pt bang cach dat an phu :
a, 3x + 14 + 5\(\sqrt{x-2}\) = 7(\(\sqrt{x+1}+\sqrt{x^2-x-2}\) )
b, 7\(\sqrt{3x-7}\) +(4x-7)\(\sqrt{7-x}\) =32
Giai pt:
\(\dfrac{6x-3}{\sqrt{x}-\sqrt{1-x}}=3+2\sqrt{x-x^2}\)
giai pt
\(\sqrt{x+5}-\sqrt{x+1}\left(\sqrt{x^2+6x+5}+1\right)=4\)
giai pt
\(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)
\(\sqrt{x^2-x+4}-x^2+x+2=0\)
\(\sqrt[3]{x+7}+\sqrt[3]{1-x}=2\)
a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)
Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)
\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)
\(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)
\(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)
\(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)
Vậy \(S=\left\{1\right\}\)
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Giai PT
d,\(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\)
Chuyển vế đặt điều kiện bình phương 2 vế là ra liền
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Xem thêm câu trả lời
giai pt
a) \(\sqrt{1+\sqrt{1-x^2}.}[\sqrt{\left(1-x\right)^3}-\sqrt{\left(1+x\right)^3}]=2+\sqrt{1-x^2}\)
b) \(\sqrt{1-x}-2x\sqrt{1-x^2}-2x^2+1=0\)
c) \(64x^6-112x^4+56x^2-7=2\sqrt{1-x^2}\)
a/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\ge0\\\sqrt{1+x}=b\ge0\end{matrix}\right.\) được hệ:
\(\left\{{}\begin{matrix}\sqrt{1+ab}\left(a^3-b^3\right)=2+ab\\a^2+b^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)\left(a^2+ab+b^2\right)=a^2+b^2+ab\\a^2+b^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1+ab}\left(a-b\right)=1\\a^2+b^2=2\end{matrix}\right.\) \(\left(a\ge b\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(a-b\right)^2=1\\a^2+b^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(1+ab\right)\left(2-2ab\right)=1\\a^2+b^2=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}1-a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2b^2=\frac{1}{2}\\a^2+b^2=2\end{matrix}\right.\)
Theo Viet đảo, \(a^2;b^2\) là nghiệm của:
\(t^2-2t+\frac{1}{2}=0\Rightarrow\left[{}\begin{matrix}t=\frac{2+\sqrt{2}}{2}\\t=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}1-x=\frac{2+\sqrt{2}}{2}\\1-x=\frac{2-\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\sqrt{2}}{2}\\x=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
2 phần còn lại ko biết giải theo kiểu lớp 10, chỉ biết lượng giác hóa, bạn tham khảo thôi :(
b/ Đặt \(x=cos2t\) pt trở thành:
\(\sqrt{1-cos2t}-2cos2t.\sqrt{1-cos^22t}-\left(2cos^22t-1\right)=0\)
\(\Leftrightarrow\sqrt{2}sint-2sin2t.cos2t-cos4t=0\)
\(\Leftrightarrow\sqrt{2}sint-sin4t-cos4t=0\)
\(\Leftrightarrow\sqrt{2}sint=sin4t+cos4t=\sqrt{2}sin\left(4t+\frac{\pi}{4}\right)\)
\(\Leftrightarrow sin\left(4t+\frac{\pi}{4}\right)=sint\)
\(\Leftrightarrow\left[{}\begin{matrix}4t+\frac{\pi}{4}=t+k2\pi\\4t+\frac{\pi}{4}=\pi-t+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\frac{\pi}{12}+\frac{k2\pi}{3}\\t=-\frac{\pi}{20}+\frac{k2\pi}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=cos\left(-\frac{\pi}{6}+\frac{k4\pi}{3}\right)\\x=cos\left(-\frac{\pi}{10}+\frac{k4\pi}{5}\right)\end{matrix}\right.\) với \(k\in Z\)
c/ Đặt \(x=cost\)
\(64cos^6t-112cos^4t+56cos^2t-7=2\sqrt{1-cos^2t}\)
\(\Leftrightarrow64cos^6t-112cos^4t+56cos^2t-7=2sint\)
Nhận thấy \(cost=0\) không phải nghiệm, pt tương đương:
\(64cos^7t-112cos^5t+56cos^3t-7cost=2sint.cost\)
\(\Leftrightarrow cos7t=sin2t=cos\left(\frac{\pi}{2}-2t\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7t=\frac{\pi}{2}-2t+k2\pi\\7t=2t-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{18}+\frac{k2\pi}{9}\\t=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=cos\left(\frac{\pi}{18}+\frac{k2\pi}{9}\right)\\x=\left(-\frac{\pi}{10}+\frac{k2\pi}{5}\right)\end{matrix}\right.\)
Ý tưởng của người ra đề khá kì quặc, công thức \(cos7a\) kia thực sự là chứng minh rất mất thời gian