(x^2-4).(x^2-9)<_0
Tính nhẩm
3 x 4 = 2 x 6 = 4 x 3 = 5 x 6 =
3 x 7 = 2 x 8 = 4 x 7 = 5 x 4 =
3 x 5 = 2 x 4 = 4 x 9 = 5 x 7 =
3 x 8 = 2 x 9 = 4 x 4 = 5 x 9 =
Học sinh nhẩm và ghi kết quả như sau:
3 x 4 = 12 2 x 6 = 12 4 x 3 = 12 5 x 6 = 30
3 x 7 = 21 2 x 8 = 16 4 x 7 = 28 5 x 4 = 20
3 x 5 = 15 2 x 4 = 8 4 x 9 = 36 5 x 7 = 35
3 x 8 = 24 2 x 9 = 18 4 x 4 = 16 5 x 9 = 45
3 x4 = 12
3 x7 = 21
3 x5 = 15
3 x8 =24
2 x 6= 12
2 x8 = 16
2 x4 =8
2 x9 =18
4 x3 =12
4 x7 =28
4 x9 =36
4 x4 =16
5 x6 =30
5 x4 =20
5 x7 =35
5 x9 =45
3 x 4 = 12 2 x 6 = 12 4 x 3 = 12 5 x 6 = 30
3 x 7 = 21 2 x 8 = 16 4 x 7 = 28 5 x 4 = 20
3 x 5 = 15 2 x 4 = 8 4 x 9 = 36 5 x 7 = 35
3 x 8 = 24 2 x 9 = 18 4 x 4 = 16 5 x 9 = 45
Tính bằng cách thuận tiện
a. 5/7 x 5/9 + 4/9 x 5/7
b. 1/10 + 5/9 + 4/9 + 9/10 – 1
c. 5/7 x 5/9 + 4/9 x 5/7 + 2/7
d. 2/7 + 2/8 + 1/4 + 1/7 + 4/7
e. 4/5 + 3/10 + 2/10 + 0,7
f. 326 x 728 + 326 x 272
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)
37 x 5^4 trên 25 ^ 2
2 ^4 x 2^6 x 3^8 x 9^2 trên 4^4 x 3^11
3 x 9^4 x 9^3 trên 3^2 x 9
125 x 5 x 64 - 25^3 x10 x 4 trên 5^7 x 8
\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)
Tính:
a) 9 x 2 + 4 = ................
= ................
b) 9 x 4 x 2 = ................
= ................
c) 9 x 9 – 18 = ................
= ................
d) 9 x 6 : 3 = ................
= ................
a) 9 x 2 + 4 = 18 + 47 = 65
b) 9 x 4 x 2 = 36 x 2 = 72
c) 9 x 9 – 18 = 81 – 18 = 63
d) 9 x 6 : 3 = 54 : 3 = 18
Tính nhẩm
9 x 1 = ..... | 9 x 2 = ..... | 9 x 3 = ..... | 9 x 4 = ..... |
1 x 9 = ..... | 2 x 9 = ..... | 3 x 9 = ..... | 4 x 9 = ..... |
9 x 5 = ..... | 9 x 6 = ..... | 9 x 7 = ..... | 9 x 8 = ..... |
5 x 9 = ..... | 6 x 9 = ..... | 7 x 9 = ..... | 8 x 9 = ..... |
9 x 1 = 9 | 9 x 2 = 18 | 9 x 3 = 27 | 9 x 4 = 36 |
1 x 9 = 9 | 2 x 9 = 18 | 3 x 9 = 27 | 4 x 9 = 36 |
9 x 5 = 45 | 9 x 6 = 54 | 9 x 7 = 63 | 9 x 8 = 72 |
5 x 9 = 45 | 6 x 9 = 54 | 7 x 9 = 63 | 8 x 9 =72 |
Bài 2: Tìm x biết:
1,x\(^2\)+4x+4=25
2,(5-2x)\(^2\)-16=0
3,(x-3)\(^3\)-(x-3)(x\(^2\)+3x+9)+9(x+1)\(^2\)=15
4,3(x+2)\(^2\)+(2x-1)\(^2\)-7(x-3)9x+3)=36
5,(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=1
6,(2x+1)\(^2\)-4(x+2)\(^2\)=9
7,(x+3)\(^{^{ }2}\)-(x-4)(x+8)=1
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
bài 1 : giải phương trình:
a. \(\sqrt{x+2\sqrt{ }x-1}=2\)
b. \(\sqrt{x^2-4x+4}=\sqrt{4x^212x+9}\)
c.\(\sqrt{x+4\sqrt{ }x-4}=2\)
d. \(\sqrt{x^2-6x+9}=2\)
e. \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
f. \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}
Tìm x: a) x^2-4=8(x-2) b) x^2-4x 4=9(x-2) c) 4x^2-12x 9=(5-x)^2
4/9 x (-13/-17) + (-2/17) x 4/9 + 2/9 x 4/17 = ?
Giải các PT:
a, \(\sqrt{x^2-6x+9}\) = 4 - x
b, \(\sqrt{x^2-9}\) + \(\sqrt{x^2-6x+9}\) = 0
c, \(\sqrt{x^2-2x+1}\) + \(\sqrt{x^2-4x+4}\) = 3
a) `sqrt(x^2-6x _9) = 4-x`
`<=> sqrt[(x-3)^2] =4-x`
`<=> |x-3| =4-x ( đk :x<=4)`
`<=> |x-3| = |4-x|`
`<=> [(x-3 =4-x),(x-3 = x-4):}`
`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`
Vậy `S = {7/2}`
b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`
`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `
`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`
`<=> x^2 -9 = x^2 -6x +9`
`<=> 6x = 9+9 =18`
`<=> x=3(t//m)`
Vậy `S={3}`
c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`
`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`
`<=> |x-1| +|x-2| =3`
xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`
`=> 1-x +2-x =3`
`=> x = 0(t//m)`
xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`
`=> x-1 +2-x =3`
`=>1=3 (vl)`
xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`
`=> x-1+x-2 =3`
`=> x=3(t//m)`
Vậy `S = {0;3}`
a: =>|x-3|=4-x
TH1: x>=3
=>4-x=x-3
=>x=7/2(nhận)
TH2: x<3
=>3-x=4-x(loại)
b: =>căn x-3(căn x+3+căn x-3)=0
=>x-3=0
=>x=3
c: =>|x-1|+|x-2|=3
Th1: x<1
=>1-x+2-x=3
=>x=0(nhận)
TH2: 1<=x<2
=>x-1+2-x=3
=>1=3(loại)
TH3: x>=2
=>x-1+x-2=3
=>x=3