1.Phân tích đa thức thành nhân tử:
a) \(x^3+\sqrt{3}x+6x^2+6\sqrt{3}x^2\)
b) \(x^4-6\sqrt{3}x+6x^3-36\sqrt{3}\)
c) \(x^5+\sqrt{3}x^5-y^5-\sqrt{3}y^5\)
Phân tích đa thức thành nhân tử (với các căn thức đã cho đều có nghĩa)
A = \(x-y-3\left(\sqrt{x}+\sqrt{y}\right)\)
B = \(x-4\sqrt{x}+4\)
C = \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
D = \(5x^2-7x\sqrt{y}+2y\)
phân tích đa thức thành nhân tử
\(x\sqrt{x}-9\)
\(x-\sqrt{x}-6\)
\(2x+5\sqrt{x}-3\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
Tìm điều kiện xác định và phân tích các đa thức sau thành nhân tử:
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(D=\sqrt{x^2+3x+2}+\sqrt{x+1}+2\sqrt{x+2}+2\)
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
\(ĐKXĐ:x;y\ge0\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\left(\sqrt{x^3}+\sqrt{x^2y}\right)-\left(\sqrt{y^3}+\sqrt{xy^2}\right)\)
\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)\)
Tìm điều kiện của x để biểu thức xác định
a) \(\sqrt{-2x^2+3}\)
b) \(\sqrt{6x^2-6}\)
c) \(\sqrt{\dfrac{3}{-x^2+5}}\)
d) \(\sqrt{-x^3-5}\)
a: ĐKXĐ: \(-\dfrac{\sqrt{6}}{2}\le x\le\dfrac{\sqrt{6}}{2}\)
b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
c: ĐKXĐ: \(-\sqrt{5}< x< \sqrt{5}\)
d: ĐKXĐ: \(x\le\sqrt[3]{-5}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
e/ ĐKXĐ: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)
Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Rightarrow5c\left(a-b\right)=5c^2\)
\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)
f/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}-2\sqrt{x+2}+2x-2\sqrt{x\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{\frac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+2}{x}}-2\sqrt{x}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)
Phân tích đa thức thành nhân tử
1. \(x\sqrt{x}+\sqrt{x}-x-1\)
2. \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
3. \(x-\sqrt{x}-2\)
4. \(x-3\sqrt{x}+2\)
5. \(-6x+5\sqrt{x}+1\)
6. \(x+4\sqrt{x}+3\)
7. \(3\sqrt{a}-2a-1\)
8. \(x+2\sqrt{x-1}\)
9. \(7\sqrt{x}-6x-2\)
10. \(x-5\sqrt{x}+6\)
11. \(x-2+\sqrt{x^2-4}\)
1) ta có : \(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)
2) ta có : \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
3) ta có : \(x-\sqrt{x}-2=x+\sqrt{x}-2\sqrt{x}-2\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)
4) ta có : \(x-3\sqrt{x}+2=x-\sqrt{x}-2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
5) ta có : \(-6x+5\sqrt{x}+1=-6x+6\sqrt{x}-\sqrt{x}+1\)
\(=6\sqrt{x}\left(1-\sqrt{x}\right)+\left(1-\sqrt{x}\right)=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)
6) ta có : \(x+4\sqrt{x}+3=x+\sqrt{x}+3\sqrt{x}+3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)\)
7) ta có : \(3\sqrt{a}-2a-1=-2a+2\sqrt{a}+\sqrt{a}-1\)
\(=-2\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)=\left(1-2\sqrt{a}\right)\left(\sqrt{a}-1\right)\)
8) ta có : \(x+2\sqrt{x-1}=x-1+2\sqrt{x-1}+1\)
\(=\left(\sqrt{x-1}+1\right)^2\)
9) ta có : \(7\sqrt{x}-6x-2=-6x+3\sqrt{x}+4\sqrt{x}-2\)
\(=-3\sqrt{x}\left(2\sqrt{x}-1\right)+2\left(2\sqrt{x}-1\right)=\left(2-3\sqrt{x}\right)\left(2\sqrt{x}-1\right)\)
10) ta có : \(x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
11) ta có : \(x-2+\sqrt{x^2-4}=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-2\right)\left(x+2\right)}\)
\(=\sqrt{x-2}\left(\sqrt{x-2}+\sqrt{x+2}\right)\)
Phân tích thành nhân tử
\(x+\sqrt{x}\)
\(x-\sqrt{x}\)
\(a+3\sqrt{a}-10\)
\(x\sqrt{x}+\sqrt{x}-x-1\)
\(x+\sqrt{x}-2\)
\(x-5\sqrt{x}+6\)
\(x\sqrt{x}-1\)
\(x\sqrt{x}-x+\sqrt{x}-1\)
\(x+2\sqrt{x}-15\)
\(x-2\sqrt{x}-3\)
\(a+\sqrt{a}-6\)
\(x-16\)
\(x+2\sqrt{x}+1\)
\(x-1\)
\(x-2\sqrt{x}+1\)
\(a\sqrt{a}+1\)
\(a+\sqrt{a}-2\)
\(2x-5\sqrt{x}+3\)
\(x-9\)
\(x+\sqrt{x}-6\)
1. $x+\sqrt{x}=\sqrt{x}(\sqrt{x}+1)$
2. $x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1)$
3. $a+3\sqrt{a}-10=(a-2\sqrt{a})+(5\sqrt{a}-10)$
$=\sqrt{a}(\sqrt{a}-2)+5(\sqrt{a}-2)=(\sqrt{a}+5)(\sqrt{a}-2)$
4. $x\sqrt{x}+\sqrt{x}-x-1=(x\sqrt{x}+\sqrt{x})-(x+1)=\sqrt{x}(x+1)-(x+1)$
$=(x+1)(\sqrt{x}-1)$
5. $x+\sqrt{x}-2=(x-\sqrt{x})+(2\sqrt{x}-2)$
$=\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)=(\sqrt{x}-1)(\sqrt{x}+2)$
6. $x-5\sqrt{x}+6=(x-2\sqrt{x})-(3\sqrt{x}-6)=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)=(\sqrt{x}-2)(\sqrt{x}-3)$
7. $x\sqrt{x}-1=(\sqrt{x})^3-1^3=(\sqrt{x}-1)(x+\sqrt{x}+1)$
8. $x\sqrt{x}-x+\sqrt{x}-1=x(\sqrt{x}-1)+(\sqrt{x}-1)=(\sqrt{x}-1)(x+1)$
9. $x+2\sqrt{x}-15=(x-3\sqrt{x})+(5\sqrt{x}-15)=\sqrt{x}(\sqrt{x}-3)+5(\sqrt{x}-3)=(\sqrt{x}-3)(\sqrt{x}+5)$
10. $x-2\sqrt{x}-3=(x+\sqrt{x})-(3\sqrt{x}+3)=\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}+1)=(\sqrt{x}+1)(\sqrt{x}-3)$
\(x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\\ x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\\ a+3\sqrt{a}-10=a+5\sqrt{a}-2\sqrt{a}-10=\sqrt{a}\left(\sqrt{a}+5\right)-2\left(\sqrt{a}+5\right)=\left(\sqrt{a}-2\right)\left(\sqrt{a}+5\right)\)
\(x\sqrt{x}+\sqrt{x}-x-1=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)=x\left(\sqrt{x}-1\right)+\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+1\right)\\ x+\sqrt{x}-2=x+2\sqrt{x}-\sqrt{x}-2=\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
Mấy bạn còn lại tương tự những bài trên nhé. Nếu còn thắc mắc ở chỗ nào bạn có thể liên hệ mình nhé. Nhớ lần sau bạn tách ra nha, chứ nhiều câu quá.
1) \(\sqrt{x^2-4x+5}+3=4x-x^2\)
2) \(4\sqrt{x^2-6+6}=x^2-6x +9\)
3) \(\sqrt{x^2-3x^3}+\sqrt{x^2-3x+6}=3\)
4) \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)