B1; Giải các phương trình sau:
a, \(2\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+1}=4\)
b,\(\sqrt{x}+\sqrt{x+1}=1+\sqrt{x\left(x+1\right)}\)
c,\(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
1. Giải các phương trình sau:
a)\(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt[]{x+\sqrt{x^2-1}}=2\)
b)\(x^2-x-\sqrt{x^2-x+13}=7\)
c)\(x^2+2\sqrt{x^2-3x+1}=3x+4\)
d)\(2x^2+5\sqrt{x^2+3x+5}=23-6x\)
e)\(\sqrt{x^2+2x}=-2x^2-4x+3\)
f)\(\sqrt{\left(x+1\right)\left(x+2\right)}=x^2+3x+4\)
2. Giải các bất phương trình sau:
1)\(\sqrt{x^2-4x+5}\ge2x^2-8x\)
2)\(2x^2+4x+3\sqrt{3-2x-x^2}>1\)
3)\(\dfrac{\sqrt{-3x+16x-5}}{x-1}\le2\)
4)\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
5)\(\dfrac{9x^2-4}{\sqrt{5x^2-1}}\le3x+2\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\) ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ; 2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)
À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)
Giải các phương trình sau: (hệ phương trình)
1.\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
2.\(\sqrt{3-4x}+\sqrt{4x+1}=-16x^2-8x+1\)
3. \(\sqrt{x^2-2x+5}+\sqrt{x+1}=2\)
4. \(\left(-4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\)
5. \(\sqrt{-4x-1}+\sqrt{4x^2+8x+3}=-4x^2-4x\)
6. \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\sqrt{\frac{x+1}{x-3}}=-3\)
7. \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
Ai làm được 4 bài hoặc nhiều hơn mik sẽ tick nha :)
Giải các phương trình sau:
1/ \(2x^2-8x+\sqrt{x^2-4x+16}=4\)
2/\(3\left(x^2+2\right)=10\sqrt{x^3+1}\)
3/\(\sqrt{3\left(1-x\right)}-\sqrt{3+x}=2\)
Thấy : \(x^2-4x+16=\left(x-2\right)^2+12>0\forall x\)
P/t \(\Leftrightarrow2\left(x^2-4x+16\right)-36+\sqrt{x^2-4x+16}=0\)
Đặt \(t=\sqrt{x^2-4x+16}>0\) ; khi đó :
\(2t^2+t-36=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-\dfrac{9}{2}\left(L\right)\end{matrix}\right.\)
Với t = 4 hay \(\sqrt{x^2-4x+16}=4\Leftrightarrow x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy ...
Câu 1 bạn trên giải rồi mik k giải nx nha
2/ \(3\left(x^2+2\right)=10\sqrt{x^3+1}\)
\(3\left(x^2-x+1\right)+3\left(x+1\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b\ge0\end{matrix}\right.\)
pt⇔ \(3a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3b=b\\a=3b\end{matrix}\right.\)
Đến đây bạn tự giải tiếp nha
3/ \(\sqrt{3-3x}-\sqrt{3+x}=2\)
\(\left(\sqrt{3-3x}-3\right)-\left(\sqrt{3+x}-1\right)=0\)
\(\dfrac{-3\left(x+2\right)}{\sqrt{3-3x}+3}-\dfrac{x+2}{\sqrt{3+x}+1}=0\)
+) \(x=-2\left(TM\right)\)
+) \(x\ne-2\Rightarrow\dfrac{-3}{\sqrt{3-3x}+3}-\dfrac{1}{\sqrt{3+x}+1}=0\)
Vì VT<0 => ptvn
2 ) ĐK : \(x\ge-1\)
P/t \(\Leftrightarrow9\left(x^2+2\right)^2=100\left(x^3+1\right)\)
\(\Leftrightarrow9x^4+36x^2+36=100x^3+100\)
\(\Leftrightarrow9x^4-100x^3+36x^2-64=0\)
\(\Leftrightarrow\left(x^2-10x-8\right)\left(9x^2-10x+8\right)=0\)
\(\Leftrightarrow x^2-10x-8=0\) ( 9x^2 - 10x + 8 > 0 )
\(\Leftrightarrow x=5\pm\sqrt{33}\) ( t/m )
Vậy ...
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
Giải các phương trình sau:
a) \(\sqrt{2x-1}+\sqrt{x-1}=5\)
b) \(x^2+2x+7=3\sqrt{\left(x^2+1\right).\left(x+3\right)}\)
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: \(x\ge1\))
\(\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{x-1}\right)^2=5^2\)
\(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow3x-2+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-1\right)}=\dfrac{27-3x}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{27-3x}{2}\ge0\\\left(2x-1\right)\left(x-1\right)=\left(\dfrac{27-3x}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}27-3x\ge0\\2x^2-2x-x+1=\dfrac{729-162x+9x^2}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\le27\\8x^2-12x+4=9x^2-162x+729\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x^2-150x+725=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x-5=0\\x-145=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=145\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=5\)
a) \(\left(x^2-5x+6\right)\left(\sqrt{x+5}+4\right)=3x^3-10x^2-7x+30\)
b) \(\sqrt{x^2+x+2}+\sqrt{x^2-x+2}=2x+1\)
c) \(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
a.
ĐKXĐ: \(x\ge-5\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(\sqrt{x+5}+4\right)=\left(3x+5\right)\left(x^2-5x+6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+6=0\\\sqrt{x+5}+4=3x+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\\sqrt{x+5}=3x+1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\x+5=9x^2+6x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\9x^2+5x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{4}{9}\end{matrix}\right.\)
b. Bạn coi lại đề, pt này nghiệm rất xấu
c.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
1) Giải phương trình
a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
b) \(x^2-8x+18=\sqrt{2x-7}+\sqrt{9-2x}\)
2) rút gọn
a) M = \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
b) N= \(\sqrt{10+\sqrt{27}+\sqrt{40}+\sqrt{60}}\)
c) C = \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
1. a) Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x+2\sqrt{x-1}}.2=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x+\sqrt{2-1}}.2=\frac{x+3}{2}\)
Bạn tự khai triển ra nha!
b) Tương tự
2) Tự làm
Ps: Ms lớp 6 nên chỉ làm được như vậy thôi! Bạn tự khai triển thành bài nhé!
1)
a) đk x>=1
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
vs x>=2
thì pt có dạng
\(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)
\(4\sqrt{x-1}=x+3\)
\(16x-16=x^2+6x+9\)
\(x^2-10x+25=0\)
x=5(tm)
vs 0<=x<1
pt \(2=\frac{x+3}{2}\)
\(x+3=4\)
\(x=1\)
b2
ý c
\(C=\sqrt{\frac{8+2\sqrt{7}}{2}}-\sqrt{\frac{8-2\sqrt{7}}{2}}=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)