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19 tháng 8 2019

ráng làm nốt rồi đi ngủ thoyy

1.

a) ĐK: \(x\ge2\)

\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}-\sqrt{x-2}-\sqrt{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)

Vậy...

b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)

Vậy...

c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)

Nhân cả 2 vế với \(\sqrt{2}\) ta được :

\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)

\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)

Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)

\(=\left|\sqrt{2x-1}+1\right|+\left|1-\sqrt{2x-1}\right|\ge\left|\sqrt{2x-1}+1+1-\sqrt{2x-1}\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)

2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)

\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(xz+yz+z^2+xy\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)

Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)

Tương tự 2 trường hợp còn lại ta đều được \(B=0\)

Vậy \(B=0\)

19 tháng 8 2019

yeu

NV
14 tháng 1 2021

1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

NV
14 tháng 1 2021

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

bach nhac lam Xl nha đến đây -----> bí

1 tháng 1 2020

Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ

Mn giúp em vs ạ! Thanks trước!

1. \(x^3-x^2+12x\sqrt{x-1}+20=0\) 2. \(x^3+\sqrt{\left(x-1\right)^3}=9x+8\) 3. \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\) 4. \(x^6+\left(x^3-3\right)^3=3x^5-9x^2-1\) 5. \(x^2-6\left(x+3\right)\sqrt{x+1}+14x+3\sqrt{x+1}+13=0\) 6. \(x^2-4x+\left(x-3\right)\sqrt{x^2-x+1}=-1\) 7. \(\sqrt{2x-1}+\sqrt{5-x}=x-2+2\sqrt{-2x^2+11x-5}\) 8. \(\sqrt{5x+11}-\sqrt{6-x}+5x^2-14x-60=0\) 9. \(x^2+6x+8=3\sqrt{x+2}\) 10. \(2x^2+3x-2=\left(2x-1\right)\sqrt{2x^2+x-3}\) 11....
Đọc tiếp

1. \(x^3-x^2+12x\sqrt{x-1}+20=0\)

2. \(x^3+\sqrt{\left(x-1\right)^3}=9x+8\)

3. \(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\)

4. \(x^6+\left(x^3-3\right)^3=3x^5-9x^2-1\)

5. \(x^2-6\left(x+3\right)\sqrt{x+1}+14x+3\sqrt{x+1}+13=0\)

6. \(x^2-4x+\left(x-3\right)\sqrt{x^2-x+1}=-1\)

7. \(\sqrt{2x-1}+\sqrt{5-x}=x-2+2\sqrt{-2x^2+11x-5}\)

8. \(\sqrt{5x+11}-\sqrt{6-x}+5x^2-14x-60=0\)

9. \(x^2+6x+8=3\sqrt{x+2}\)

10. \(2x^2+3x-2=\left(2x-1\right)\sqrt{2x^2+x-3}\)

11. \(\sqrt{x+1}+\sqrt{4-x}-\sqrt{\left(x+1\right)\left(4-x\right)}=1\)

12. \(x^2-\sqrt{x^2-4x}=4\left(x+3\right)\)

13. \(x^2-x-4=2\sqrt{x-1}\left(1-x\right)\)

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\)

15. \(\sqrt{2x^2+3x+2}+\sqrt{4x^2+6x+21}=11\)

16. \(\sqrt{x+3+3\sqrt{2x-3}}+\sqrt{x-1+\sqrt{2x-1}}=2\sqrt{2}\)

17. \(\left(x-2\right)^2\left(x-1\right)\left(x-3\right)=12\)

18. \(2x^2+\sqrt{x^2-2x-19}=4x+74\)

19. \(x^4+x^2-20=0\)

20. \(x+\sqrt{4-x^2}=2+3x\sqrt{4-x^2}\)

21. \(\left(x^2+x+1\right)\left(\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1\right)=9\)

22. \(\sqrt{x^2-3x+5}+x^2=3x+7\)

23. \(x^2+6x+5=\sqrt{x+7}\)

24. \(\frac{2x^2-3x+10}{x+2}=3\sqrt{\frac{x^2-2x+4}{x+2}}\)

25. \(5\sqrt{x-1}-\sqrt{x+7}=3x-4\)

26. \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)

27. \(\sqrt{x-1}+\sqrt{5-x}-2=2\sqrt{\left(x-1\right)\left(5-x\right)}\)

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\)

29. \(\frac{26x+5}{\sqrt{x^2+30}}+2\sqrt{26x+5}=3\sqrt{x^2+30}\)

30. \(\frac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\frac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)

12
20 tháng 3 2020

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)

PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)

Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)

giai tiep

20 tháng 3 2020

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{matrix}\right.\)

6 tháng 8 2018

Vãi cả "Toán Lớp 1"

26 tháng 2 2022

đây đích thực có phải lớp 1 ko ak?

chắc bn đây phải cấp 2 r