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AH
Akai Haruma
Giáo viên
13 tháng 8 2021

Bài đã đăng rồi bạn lưu ý không đăng lại làm loãng box toán.

31 tháng 8 2015

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x2 - 16x - 34 = 10x2 + 3x - 34

=> 10x2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

2 tháng 1 2016

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

5 tháng 7 2018

1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

5 tháng 7 2018

2) a)\(x\left(x+3\right)-x^2=6\)

\(\Rightarrow x^2+3x-x^2=6\)

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

c) x(x+5)-(x+1)(x-2)=7

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

NV
13 tháng 8 2021

Bài này sử dụng tính chất cơ bản: \(\left|A\right|\pm A\ge0\) với mọi A

a.

\(A=\left|-x-3\right|+\left|4x+1\right|+\left|3x+5\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|3x+5\right|+5x+2=\left|3x-2\right|+\dfrac{3}{2}.\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|2x+\dfrac{10}{3}\right|+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|5x+\dfrac{4}{3}\right|+5x+\dfrac{4}{3}+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+\dfrac{2}{3}\ge\dfrac{2}{3}\)

\(A_{min}=\dfrac{2}{3}\) khi \(2x+\dfrac{10}{3}=0\Rightarrow x=-\dfrac{5}{3}\)

b. Tương tự

\(B\ge\left|5x+7\right|+\left|x+\dfrac{5}{4}\right|+3\left|x+\dfrac{5}{4}\right|-6x+5\)

\(B\ge\left|6x+\dfrac{33}{4}\right|-\left(6x+\dfrac{33}{4}\right)+3\left|x+\dfrac{5}{4}\right|+\dfrac{53}{4}\ge\dfrac{53}{4}\)

\(B_{min}=\dfrac{53}{4}\) khi \(x=-\dfrac{5}{4}\)

AH
Akai Haruma
Giáo viên
13 tháng 8 2021

Lời giải:

a. Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:

\(A=|-x-3|+|4x+1|+|3x+5|+5x+2\)

\(\geq |-x-3+4x+1|+|3x+5|+5x+2=|3x-2|+|3x+5|+5x+2\)

Nếu $x\geq \frac{2}{3}$ thì:

$A\geq 3x-2+3x+5+5x+2=11x+5\geq 11.\frac{2}{3}+5=\frac{37}{3}$

Nếu $\frac{-5}{3}\leq x< \frac{2}{3}$ thì:

$A\geq 2-3x+3x+5+5x+2=9+5x\geq 9+5.\frac{-5}{3}=\frac{2}{3}$

Nếu $x< \frac{-5}{3}$ thì:

$A\geq 2-3x-3x-5+5x+2=-1-x>\frac{2}{3}$

Từ 3 TH trên suy ra $A_{\min}=\frac{2}{3}$ khi $x=\frac{-5}{3}$

5 tháng 9 2019

a) 3x(4x - 3) - 2x(5 - 6x) = 0

=> 6x2 - 9x - 10x + 12x2 = 0

=> 18x2 - 19x = 0

=> x(18x - 19) = 0

=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)

b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0

=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0

=> 8x - 15 = 0

=> 8x = 15

=> x = 15 : 8 = 15/8

c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)

=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x

=> 4x - x2 - 5x2 - 15x = 0

=> -6x2 - 11x = 0

=> -x(6x - 11) = 0

=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)

5 tháng 9 2019

a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)

b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)

25 tháng 8 2020

a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7

<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7

<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7

<=> -8x - 4 = 7

<=> -8x = 11

<=> x = -11/8

b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8

<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8

<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8

<=> 27x + 21 = 8

<=> 27x = -13

<=> x = -13/27 

c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9

<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9

<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9

<=> 3x - 6 = 9

<=> 3x = 15

<=> x = 5

d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10

<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10

<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10

<=> -11x + 6 = 10

<=> -11x = 4

<=> x = -4/11

25 tháng 8 2020

a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)

b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)

c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)

d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)

28 tháng 1 2021

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Leftrightarrow156-56x=24x-324\)

\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)

\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)

\(\Leftrightarrow7x+37=11x-35\)

\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)

28 tháng 1 2021

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)

\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)

\(\Leftrightarrow5x+33x-18-182=0\)

\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)