1 /2 *x^2 -19 /6 *x+1
Bài 1 tính hợp lí
a) 2017.;1-2018)+2018.2017+(-1)*2017
b) 7/19 8/11+3/11:19/7-2/-19
c) (6 và 4/9+3 và 7/11)-5 và4/9
Bài 2 tìm x
a) 5/6-x=-5/12-1/-4
b) x-2 trên phần -5=6/7
c) 1/2.x-3/4.x=-5/6
1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)
c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)
từ rồi làm tiếp
2)
a)\(\frac{5}{6}-x=-\frac{5}{12}-\frac{1}{-4}=-\frac{5}{12}+\frac{1}{4}=-\frac{5}{12}+\frac{3}{12}=-\frac{1}{6}.\)
\(x=\frac{5}{6}--\frac{1}{6}=\frac{5}{6}+\frac{1}{6}=1\)
b)\(\frac{x-2}{-5}=\frac{6}{7}\)
\(\frac{\left(x-2\right)\left(-7\right)}{35}=\frac{30}{35}\)
\(\Rightarrow\left(x-2\right)\left(-7\right)=30\)
c)\(\frac{1}{2}x-\frac{3}{4}x=-\frac{5}{6}\)
\(\left(\frac{1}{2}-\frac{3}{4}\right)x=-\frac{5}{6}\)
\(-\frac{1}{4}x=-\frac{5}{6}\)
\(x=\frac{20}{6}=\frac{10}{3}\)
vậy \(x=\frac{10}{3}\)
Đố bn viết thuộc bảng nhân 19 :
19 x 1 =
19 x 2 =
19 x 3 =
19 x 4 =
19 x 5 =
19 x 6 =
19 x 7 =
19 x 8 =
19 x 9 =
19 x 10 =
19 x 1 = 19
19 x 2 = 38
19 x 3 = 57
19 x 4 = 76
19 x 5 = 95
19 x 6 = 114
19 x 7 = 133
19 x 8 = 152
19 x 9 = 172
19 x 10 = 190
Hok tốt ah
MẸO HEN
NẾU DÒNG THẲNG TẮP NHƯ VẦY
9 x 1
9 x 2
9 x 3
9 x 4
9 x 5
9 x 6
9 x 7 6 3
9 x 8 7 2
9 x 9 8 1
9 x 1=9 0
Tim x x(x+5)(x-5) - (x+2)(x^2-2x+4)=5
(x+1)^3 - (x-1)^3 -6(x-1)^2 = -19
`#3107.101107`
\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)
`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`
`<=> x^3 - 25x - x^3 - 8 = 5`
`<=> -25x - 8 = 5`
`<=> -25x = 13`
`<=> x = -13/25`
Vậy, `x = -13/25`
_____
\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)
`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`
`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`
`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`
`<=> 12x - 4 = -19`
`<=> 12x = -15`
`<=> x = -15/12 = -5/4`
Vậy, `x = -5/4.`
________
`@` Sử dụng các hđt:
`1)` `A^2 + B^2 = (A - B)(A + B)`
`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`
`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`
`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`
`5)` `(A - B)^2 = A^2 - 2AB + B^2.`
a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)
=>\(x\left(x^2-25\right)-x^3-8=5\)
=>\(x^3-25x-x^3-8=5\)
=>-25x=13
=>\(x=-\dfrac{13}{25}\)
b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)
=>\(6x^2+2-6x^2+12x-6=-19\)
=>12x-4=-19
=>12x=-15
=>x=-5/4
tìm x, biết :
d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19
d. (x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1
<=> x3 - 9 + (x2 + 2x)(2 - x) = 1
<=> x3 - 9 + 2x2 - x3 + 4x - 2x2 = 1
<=> 4x = 10
<=> x = \(\dfrac{10}{4}=\dfrac{5}{2}\)
d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
\(<=> x^3-27-x(x^2-4)=1\)
\(<=> x^3-27-x^3-4x=1<=>-4x=28<=> x=-7\)
=> ptrình có tập nghiệm S={-7}
e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19
\(<=> x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+19=0\)
\(<=>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(<=>12x=15<=>x=12/15 \)
=> ptrình có tập nghiệm S={12/15}
tìm x
5 mũ x -2 ( số 2 là số mũ nhé) - 3 mũ 2 = 2 mũ 4 - ( 6 mũ 8 : 6 mũ 6 - 6 mũ 2)
(x mũ 2 - 1) mũ 4 = 81
3 mũ x + 4 mũ 2 = 19 mũ 6 : ( 19 mũ 3 * 19 mũ 2) - 3 * 1 mũ 2005( 1 * 2005 lần chứ không phải 3* 1 mũ 2005 lần đâu nhé)
mk viết khó hiểu mong các bạn thông cảm và giúp đỡ mình nhé
Đề bài có phải là như vậy ko bạn:
5x-2 - 32= 24- (68 : 66 - 62) ( x2-1)4 =81
( 3x+4 )2 = 196 : ( 193 x 192 ) - 31x 2005
Bạn xem hộ mik như vậy có đúng đề ko, đúng thì mik làm cho nhé!
a,5^x+4-3×5^x+3=2×5^11
b,3×5^x+2+4×5^x-3=19×5^10
c,6×8^x-1+8^x-1=6×8^19+8^21
d,5×2^x+3×2^x+2=5×2^5+3×2^7
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
\(b) 5^{x+4} - 3 . 5^{x+3} = 2. 5^{11} \)
\(c) 2 . 3^{x+2} + 4 . 3^{x+1} = 10 . 3 ^{6}\)
\(d) 6 . 8 ^ {x-1} + 8 ^ {x+1} = 6 . 8^ {19} + 8 ^ {21}\)
b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20
A= 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ... + 1/19 x 20
A=1/2-1/3+1/3-1/4+...+1/19-1/20
=1/2-1/20
=9/20
A= 1/2 x 3 + 1/3 x 4 + 1/4 x 5 + 1/5 x 6 + 1/6 x 7 + ... + 1/19 x 20
--> A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 +...+ 1/19 - 1/20
= 1/2 - 1/20 = 10/20 - 1/20 = 9/20
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{10}{20}-\dfrac{1}{20}=\dfrac{9}{20}\)
e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19.
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\\ \Leftrightarrow12x=-15\\ \Leftrightarrow x=-\dfrac{15}{12}=-\dfrac{5}{4}\)
\(\Leftrightarrow\)\(x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-(6x^2-12x+6)+19=0\)
\(\Leftrightarrow\)\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(\Leftrightarrow\)\(12x+15=0\)
\(\Leftrightarrow\)\(x=-\dfrac{5}{4}\)