x mũ 2 - x - (2+4+6+...+2500)=0
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a) x(x+1)=2+4+6+8+...+2500
x(x+1)=(2500+2).1250:2
x(x+1)=1563750
Ta thấy: x và x+1 là 2 số liên tiếp
Mà x(x+1)=1563750=>x=1250;x+1=1251
Vậy x=1250
b) x-3=12:{390:[500-(5^3+7^2.5)]}
x-3=12:[390:(500-254.64)]
x-3=12:(390:245.35)
x-3=12:1.58
x-3=7.54
=>x=10.5
Vậy x=10.5
c) x+3^3.12=3^3.19
=>x=2.46
Vậy x=2.46
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1.(x -5)^2 - 25 =0
=> (x - 5)^2 = 25
=> x - 5 = 5 hoặc x - 5 = -5
=> x = 10 hoặc x = 0
vậy_
2. (x -2)^3 =27
=> x - 2 = 3
=> x = 5
vậy_
3. 3(x -7) + 2x(x+2) = 2x^2
=> 3x - 21 + 2x^2 + 4x = 2x^2
=> 7x - 21 = 0
=> 7x = 21
=> x = 3
vậy_
4. (x^2 - 4) (x +8) =0
=> x^2 - 4 = 0 hoặc x + 8 = 0
=> x^2 = 4 hoặc x = -8
=> x = 2 hoặc x = -2 hoặc x = -8
vậy_
5. x^ 2 + 3x = 0
=> x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3
vậy_
6. 3x^3 - 3x = 0
=> 3x(x^2 - 1) = 0
=> 3x(x - 1)(x + 1) = 0
=> x = 0 hoặc x = 1 hoặc x = -1
vậy_
7. (x +1)^2 = ( 2x +3)^2
=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0
=> (3x + 3)(-x - 2) = 0
=> x = -1 hoặc x = -2
vậy_
Bài làm
1) ( x - 5 )2 - 25 = 0
<=> ( x - 5 - 5 )( x - 5 + 5 ) = 0
<=> x( x - 10 ) =
<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy S = { 0; 10 }
2) \(\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm phương trình.
3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)
\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=\frac{21}{7}=3\)
Vậy x = 3 là nghiệm phương trình
4) \(\left(x^2-4\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)
Vậy S = { 2; -2; -8 }
5) \(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)
Vậy S = { 0; -3 }
6) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
Vậy S = { +1; 0 }
7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)
Vậy S = { -2; -4/3 }
# Học tốt #
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Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-5x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(x-5\right)^2=0\Leftrightarrow x=5\)
5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)
7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)
\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)
\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\left(x-4\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow x-4=0\) hoặc \(\left(x-1\right)^2=0\)
\(\Leftrightarrow x=4\) hoặc \(x-1=0\)
\(\Leftrightarrow x=4\) hoặc x=1
\(\left(2x-6\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-6=0\) hoặc \(x^2+1=0\)(vô lí)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Xét VT ta có:1+2+3+4+...+x=\(\frac{x\left(x+1\right)}{2}\)
Mà VP=78
\(\Rightarrow\frac{x\left(x+1\right)}{2}=78\)
\(\Rightarrow x\left(x+1\right)=156\)
\(x\left(x+1\right)=12\cdot13\)
\(\Rightarrow x=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ok tả lời
(57 + 75) . (68 + 86) . ( 24 - 42)
(57 + 75) . (68 + 86) . 0 = 0
HOK TỐT NHA BN
cái cuối nếu giải thẳng ra phải là
(57 + 75) . (68 + 86) . (24 - 42)
(57 + 75) . (68 + 86) .( 16 -16)
(57 + 75) . (68 + 86) . 0 = 0
HOK TỐT NHA
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. | x + 1| + (y + 2)2 = 0
Mà (y + 2)2 \(\ge\) 0
Đẳng thức khi . y + 2 \(\ge\) 0
y \(\ge\) - 2
. x + 1 = 0
. x = -1