\(\sqrt{8x^2-16x+10}+\sqrt{2x^2-4x+4}=\sqrt{7-x^2+2x}\)
Những câu hỏi liên quan
giải phương trình
a)\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
b)\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
c)\(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
d)\(\dfrac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
Đúng 0
Bình luận (0)
Tìm điều kiện có nghĩa:1) sqrt{2x^2}2) sqrt{-x}3) sqrt{-x^2-3}4) sqrt{x^2+2x+3}5) sqrt{-a^2+8a-16}6) sqrt[]{16x^2-25}7) sqrt{4x^2-49}8) sqrt{8-x^2}9) sqrt{x^2-12}10) sqrt{x^2+2x-3}11) sqrt{2x^2+5x+3}12) sqrt{dfrac{4}{x-1}}13) sqrt{dfrac{-1}{x-3}}14) sqrt{dfrac{-3}{x+2}}15) sqrt{dfrac{1}{2a-1}}16) sqrt{dfrac{2}{3-2a}}17) sqrt{dfrac{-1}{2a-5}}18) sqrt{dfrac{-2}{3-5a}}19) sqrt{dfrac{-a}{5}}20) dfrac{1}{sqrt{-3a}}
Đọc tiếp
Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
Đúng 0
Bình luận (0)
Giải các phương trình sau: (hệ phương trình)1.sqrt{3x^2+6x+7}+sqrt{5x^2+10x+14}4-2x-x^22.sqrt{3-4x}+sqrt{4x+1}-16x^2-8x+13. sqrt{x^2-2x+5}+sqrt{x+1}24. left(-4x-1right)sqrt{x^2+1}2x^2+2x+15. sqrt{-4x-1}+sqrt{4x^2+8x+3}-4x^2-4x6. left(x-3right)left(x+1right)+4left(x-3right)sqrt{frac{x+1}{x-3}}-37. sqrt{xleft(x-1right)}+sqrt{xleft(x+2right)}2sqrt{x^2}Ai làm được 4 bài hoặc nhiều hơn mik sẽ tick nha :)
Đọc tiếp
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 :)
a) \(\sqrt{3x-4}+\sqrt{4x+1}=-16x^2-8x+1\)
b) \(\sqrt{x}+2\sqrt{x+3}=7-\sqrt{x^2+3}\)
c) \(x^2-6x+26=6\sqrt{2x+1}\)
d)\(\sqrt{2006x^2-2005}+\sqrt{2005x^2-2004}=\sqrt{2006^2+2x-2003}+\sqrt{2005x^2+x-2002}\)
a) \(\sqrt{3x-4}\) + \(\sqrt{4x+1}\) = \(-16x^2 - 8x +1\) với
ĐKXĐ :
- Vế trái \(x \ge \frac{4}{3}\)
- Vế phải : \(-16x^2 - 8x +1\) \(\ge 0\) \(\Leftrightarrow \) \(x \le \frac{\sqrt{2}-1}{4}\) hoặc \(x \le \frac{-\sqrt{2}-1}{4}\)
Hai điều kiện trái ngược nhau
Vậy phương trình vô nghiệm .
Đúng 0
Bình luận (1)
Gpt :
1) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
2) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+s}+\sqrt{x+1}=16\)
3)\(\sqrt{4x+20}+\sqrt{x+5}-\frac{1}{3}\sqrt{9x+45}=4\)
4) \(\frac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải pt , sqrt{x^4+4x^2}+sqrt{x+x^2}sqrt{left(x^2+sqrt{x}right)^2+9x^2}.x0x^30x^32.0.sqrt{0}x^32xsqrt{x}x^32xsqrt{x} 4left(x^3-2xsqrt{x}right)^204left(x^6-4x^4sqrt{x}+4x^2xright)04x^6-16x^4sqrt{x}+16x^2x04x^6+16x^316x^4sqrt{x}16x^4+4x^5+4x^6+16x^316x^4+4x^5+16x^4sqrt{x}4x^3left(x+1right)left(x^2+4right)4left(4x^4+4x^4sqrt{x}+x^4.xright)4x^3left(x+1right)left(x^2+4right)4left(2x^2+x^2sqrt{x}right)^22sqrt{2x^3left(x+1right)left(x^2+4right)}2left(2x^2+x^2sqrt{x}right)x^4+x^2+4x^2+x+2sqrt{2x^3le...
Đọc tiếp
giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)
\(x=0\)
\(x^3=0\)
\(x^3=2.0.\sqrt{0}\)
\(x^3=2x\sqrt{x}\)
\(x^3=2x\sqrt{x}\)
\(4\left(x^3-2x\sqrt{x}\right)^2=0\)
\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)
\(4x^6-16x^4\sqrt{x}+16x^2x=0\)
\(4x^6+16x^3=16x^4\sqrt{x}\)
\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)
\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)
\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)
\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)
\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)
vậy x=0 là nghiệm của pt =))
cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?
Đúng 0
Bình luận (0)
Giải các phương trình sau:
a) \(2x^2+4x=5\sqrt{x^3-x^2-4}\)
b) \(2x^2-x+1=\sqrt{4x^4+1}\)
c) \(8x^2+20x+1=\sqrt{64x^4+1}\)
d) \(3x^2-4x+23=3\sqrt{x^4-8x+63}\)
e) \(4x^2-4x+1=2\sqrt{16x^4+4x+1}\)
a/ ĐKXĐ: ....
\(\Leftrightarrow2x^2+2x+4+2x-4=5\sqrt{\left(x-2\right)\left(x^2+x+2\right)}\)
\(\Leftrightarrow2\left(x^2+x+2\right)+2\left(x-2\right)=5\sqrt{\left(x-2\right)\left(x^2+x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a\\\sqrt{x-2}=b\end{matrix}\right.\)
\(\Leftrightarrow2a^2+2b^2=5ab\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\sqrt{x-2}\\2\sqrt{x^2+x+2}=\sqrt{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\left(x-2\right)\\4\left(x^2+x+2\right)=x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+10=0\\4x^2+3x+10=0\end{matrix}\right.\)
Phương trình vô nghiệm
Đúng 0
Bình luận (0)
b/ ĐKXĐ: ....
\(\Leftrightarrow2x^2-x+1=\sqrt{4x^4+4x^2+1-4x^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2+1\right)^2-\left(2x\right)^2}\)
\(\Leftrightarrow2x^2-x+1=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
\(\Leftrightarrow\frac{3}{4}\left(2x^2-2x+1\right)+\frac{1}{4}\left(2x^2+2x+1\right)=\sqrt{\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2-2x+1}=a\\\sqrt{2x^2+2x+1}=b\end{matrix}\right.\)
\(\Leftrightarrow3a^2+b^2=4ab\Leftrightarrow3a^2-4ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\\3\sqrt{2x^2-2x+1}=\sqrt{2x^2+2x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-2x+1=2x^2+2x+1\\9\left(2x^2-2x+1\right)=2x^2+2x+1\end{matrix}\right.\)
Đúng 0
Bình luận (0)
c/
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2\right)^2+16x^2+1-16x^2}\)
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2+1\right)^2-16x^2}\)
\(\Leftrightarrow8x^2+20x+1=\sqrt{\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)}\)
\(\Leftrightarrow3\left(8x^2+4x+1\right)-2\left(8x^2-4x+1\right)=\sqrt{\left(8x^2+4x+1\right)\left(8x^2-4x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{8x^2+4x+1}=a>0\\\sqrt{8x^2-4x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow3a^2-2b^2=ab\Leftrightarrow3a^2-ab-2b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a+2b\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow\sqrt{8x^2+4x+1}=\sqrt{8x^2-4x+1}\)
\(\Leftrightarrow8x^2+4x+1=8x^2-4x+1\)
\(\Leftrightarrow x=0\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
28 tháng 11 2019 lúc 23:30
Giải các pt sau:
a) sqrt{x+8}+frac{9x}{sqrt{x+8}}-6sqrt{x}0
b) x^4-2x^3+sqrt{2x^3+x^2+2}-20
c) 3xsqrt[3]{x+7}left(x+sqrt[3]{x+7}right)7x^3+12x^2+5x-6
d) 4x^2+left(8x-4right)sqrt{x}-13x+2sqrt{2x^2+5x-3}
e) 16x^2+19x+7+4sqrt{-3x^2+5x+2}left(8x+2right)left(sqrt{2-x}+2sqrt{3x+1}right)
f) left(5x+8right)sqrt{2x-1}+7xsqrt{x+3}9x+8-left(x+26right)sqrt{x-1}
g) sqrt[3]{3x+1}+sqrt[3]{5-x}+sqrt[3]{2x-9}-sqrt[3]{4x-3}0
Đọc tiếp
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!