√(60-24x-5x2)=x2+5x-10
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\(PT\Leftrightarrow-5x^2-24x+60=\left(x^2+5x-10\right)^2\\ \Leftrightarrow-5x^2-24x+60=x^4+10x^3+5x^2-100x+100\\ \Leftrightarrow x^4+10x^3+10x^2-76x+40=0\\ \Leftrightarrow x^4+4x^3-10x^2+6x^3+24x^2-60x-4x^2-16x+40=0\\ \Leftrightarrow\left(x^2+4x-10\right)\left(x^2+6x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+4x-10=0\\x^2+6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt{14}\\x=-2-\sqrt{14}\\x=-3+\sqrt{13}\\x=-3-\sqrt{13}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=3\left(x^2-8x+16\right)=3\left(x-4\right)^2\\ b,=5\left(x^2-1\right)=5\left(x-1\right)\left(x+1\right)\\ c,=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`# \text {04th5}`
`a.`
`P = (5x^2 - 2xy + y^2) - (x^2 + y^2) - (4x^2 - 5xy + 1)`
`= 5x^2 - 2xy + y^2 - x^2 - y^2 - 4x^2 + 5xy - 1`
`= (5x^2 - x^2 - 4x^2) + (-2xy + 5xy) + (y^2 - y^2) - 1`
`= 3xy - 1`
`b.`
\((x^2-5x+4)(2x+3)-(2x^2-x-10)(x-3)\)
`= x^2(2x + 3) - 5x(2x + 3) + 4(2x + 3) - [ 2x^2(x - 3) - x(x - 3) - 10(x - 3)]`
`= 2x^3 + 3x^2 - 10x^2 - 15x + 8x + 12 - (2x^3 - 6x^2 - x^2 + 3x - 19x + 30)`
`= 2x^3 -7x^2 - 7x + 12 - (2x^3 - 7x^2 - 7x + 30)`
`= 2x^3 - 7x^2 - 7x + 12 - 2x^3 + 7x^2 + 7x -30`
`= -30`
Vậy, giá trị của biểu thức không phụ thuộc vào giá trị của biến.
![](https://rs.olm.vn/images/avt/0.png?1311)
1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{5x_1-x_2}{x_1}+\dfrac{5x_2-x_1}{x_2}\)
\(=\dfrac{5x_1\cdot x_2-x_2^2+5x_1x_2-x_1^2}{x_1x_2}\)
\(=\dfrac{10x_1x_2-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{x_1x_2}\)
\(=\dfrac{10\cdot4-\left[5^2-2\cdot4\right]}{4}=\dfrac{40-25+8}{4}=\dfrac{23}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu ol thì tham khảo nah nguoiemtinhthong.
1.1
2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1
⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)
Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0
pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0
a=2ba=2b v a=13ba=13b
Các bạn tự giải quyết tiếp nhé.
1.2
TXĐ D=[1;+∞)D=[1;+∞)
đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0
pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0
⇔a=b⇔a=b v a=23ba=23b
...
1.3
D=[3;+∞)D=[3;+∞)
Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0
pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2
⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0
⇒a=5b⇒a=5b
...
1.4
ĐK
⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)
⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)
Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)
⇔2a2+2b2=3ab
1.5
Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)
⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x
⇔t2−t−4x2+2x=0t2−t−4x2+2x=0
Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2
⇒t=1−2xt=1−2x hoặc t=2xt=2x
1.1
2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1
2(.2+x+1)+3(x-1)
3a+b=11a2-19b2
tóm tắt
![](https://rs.olm.vn/images/avt/0.png?1311)
1:=x^3-27-x^2-3=x^3-x^2-30
2: =x-2+125x^3+150x^2+60x+8
=125x^3+150x^2+61x+6
3: \(=2xy-5y+5y=2xy\)
4: =25x-10x^2+15x
=-10x^2+40x
x=-căn bậc hai(13)-3, x=căn bậc hai(14)-2