a,\(8x^3-12x^2 6x-5=0\Leftrightarrow8\left(x^3-\frac{3}{2}x^2 \frac{3}{4}x-\frac{1}{8}\right)-4=0\)\(\Leftrightarrow8\left(x-\frac{1}{2}\right)^3=4\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{...

T

Tuấn

30 tháng 7 2016

a,\(8x^3-12x^2+6x-5=0\Leftrightarrow8\left(x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\right)-4=0\)
\(\Leftrightarrow8\left(x-\frac{1}{2}\right)^3=4\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{2}\Leftrightarrow x=\frac{1}{\sqrt[3]{2}}+\frac{1}{2}\)

#Toán lớp 1

2

T

Tuấn

30 tháng 7 2016

b, 3x^3+3x^2+3x+1=0<=>2x^3+(x+1)^3=0<=> .
Hằng đẳng thức đi bác

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KN

ka nekk

26 tháng 2 2022

đây đích thực có phải lớp 1 ko bn?

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RN

Ren Nishiyama

9 tháng 4 2020

1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12 \(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\) \(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6 \(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6 \(\Leftrightarrow\) 11x = 9 \(\Leftrightarrow\) x = 0,8 Vậy S = {0,8} 2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12 \(\Leftrightarrow\)...

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1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)

\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6

\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6

\(\Leftrightarrow\) 11x = 9

\(\Leftrightarrow\) x = 0,8

Vậy S = {0,8}

2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12

\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)

\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x

\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20

\(\Leftrightarrow\) 17x = 7

\(\Leftrightarrow\) x = \(\frac{7}{17}\)

Vậy S = {\(\frac{7}{17}\)}

3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15

\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)

\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3

\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3

\(\Leftrightarrow\) 4x = 2

\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)

Vậy S = {\(\frac{1}{2}\)}

4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12

\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)

\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24

\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24

\(\Leftrightarrow\) 5x = -58

\(\Leftrightarrow\) x = -11,6

Vậy S = {-11,6}

5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)

\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x

\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10

\(\Leftrightarrow\) -8x = -1

\(\Leftrightarrow\) x = \(\frac{1}{8}\)

Vậy S = {\(\frac{1}{8}\)}

6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12

\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)

\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x

\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3

\(\Leftrightarrow\) -5x = -2

\(\Leftrightarrow x=\frac{2}{5}\)

7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)

\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0

\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4

\(\Leftrightarrow\) -3x = -18

\(\Leftrightarrow\) x = 6

Vậy S = {6}

8) x\(^2\) - x - 6 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0

\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0

\(\Leftrightarrow\) (x - 3).(x + 2) = 0

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -2

Vậy S = {3; -2}

#Toán lớp 8

0

LT

Lê Tài Bảo Châu

8 tháng 2 2020

bt em gửi cô...

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#Toán lớp 8

5

K

KAl(SO4)2·12H2O

8 tháng 2 2020

Bài 1 dài dòng quá em :( Rút gọn bớt cũng được thì phải

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LT

Lê Tài Bảo Châu

8 tháng 2 2020

Chị ơi bài 1 em sai cái gì ko ạ ? đk x khác 3 mà đúng ko

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RN

Ren Nishiyama

19 tháng 4 2020

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\) \(\Leftrightarrow x^2+5x-2x-10-5=1x+5\) \(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0 \(\Leftrightarrow\) \(x^2+2x-20=0\) \(\Leftrightarrow x^2+2x-10x-20=0\) \(\Leftrightarrow\) (x\(^2\) + 2x) - (10x +...

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3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

\(\Leftrightarrow\) \(x^2+2x-20=0\)

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4

\(\Leftrightarrow\) 4x - 4 = 0

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

#Toán lớp 8

0

NL

Nguyen Le Minh Thu

18 tháng 8 2020

Giải phương trình:1) \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)2) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)3)\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)4)\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)5)\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)6)\(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)7)\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)8)\(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)giải giúp mik với...

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#Toán lớp 8

2

KN

Khánh Ngọc

18 tháng 8 2020

1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow95x-5=96-6x\)

\(\Leftrightarrow95x+6x=96+5\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9=32x+60\)

\(\Leftrightarrow30x-32x=60-9\)

\(\Leftrightarrow-2x=51\)

\(\Leftrightarrow x=-\frac{51}{2}\)

3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)

\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)

\(\Leftrightarrow2x+1=5x+1\)

\(\Leftrightarrow2x=5x\)

\(\Leftrightarrow x=0\)

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HQ

Huỳnh Quang Sang

19 tháng 8 2020

4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)

=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)

=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)

=> 27 - 9x + 80 - 16x = 12 - 12x - 48

=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0

=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0

=> 143 - 13x = 0

=> 13x = 143

=> x = 11

5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)

=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)

=> 6x - 18 + 7x - 35 - 13x - 4 = 0

=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0

=> -57 = 0(vô nghiệm)

6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)

=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)

=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)

=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)

=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)

=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)

=> \(12x+10-10x-3=12x+2\)

=> \(2x+10-3=12x+2\)

=> 2x + 10 - 3 - 12x - 2 = 0

=> (2x - 12x) + (10 - 3 - 2) = 0

=> -10x + 5 = 0

=> -10x = -5

=> x = 1/2

7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)

=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)

=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)

=> 6x - 3 - 5x + 10 - x - 7 = 0

=> (6x - 5x - x) + (-3 + 10 - 7) = 0

=> 0x + 0 = 0

=> 0x = 0

=> x tùy ý

Bài 8 tự làm nhé

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YN

Yến Nhi

24 tháng 2 2020

Giải các phương trình...

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#Toán lớp 8

1

YN

Yến Nhi

25 tháng 2 2020

giup minh voi cac bạn

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TP

Thị Phương Thảo Trần

24 tháng 6 2020

1. tinh` giá trị biểu thức ( tính nhanh nếu có thế ) \(a)\frac{-6}{11}.\frac{5}{13}+\frac{-6}{11}.\frac{8}{13}-\left(\frac{-2}{5}\right)^0\) \(b)\left(2\frac{2}{3}+3\frac{1}{2}\right);\left(4\frac{3}{4}-2\frac{1}{6}\right)+\frac{19}{31}\) \(c)2,4:\left(-2\right)^3+\left(3-\frac{9}{11}\right).1\frac{3}{8}\) \(d)\left(-\frac{3}{4}\right)^2:\frac{-3}{8}+\frac{1}{2}-\frac{3}{4}-\left(\frac{-78}{57}\right)^0\) 2. tìm x \(a)x+\frac{-1}{5}=\left(-\frac{3}{4^{ }}\right)^2\)...

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#Toán lớp 6

0

NG

Não Gà

24 tháng 6 2020

bạn tự làm đi tính toán thôi mà

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HN

hội những fan của Noo

13 tháng 1 2018

a)\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)

b)\(\left(\frac{4}{5}+x\right).\left(x-\frac{8}{13}\right)=0\)

c)\(\left(2x-\frac{1}{2}\right).\left(x-3\right)=0\)

d)\(x+3\frac{1}{2}x+x=\frac{1}{2}\)

#Toán lớp 7

1

Q

QuocDat

13 tháng 1 2018

a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)

Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)

b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)

Vậy x=-4/5 hoặc x=8/13

c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)

Vậy x=1/4 hoặc x=3

\(x+\frac{7}{2}x+x=\frac{1}{2}\)

\(2x+\frac{7}{2}x=\frac{1}{2}\)

\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)

\(\frac{11}{2}x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{11}{2}\)

\(x=\frac{1}{11}\)

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