a) 17x + 33x = 100

<=> 50x = 100

<=> x = 2

Vậy x = 2

b) 23x + 15 ( x + 7) = 105

<=> 23x + 15x + 105 = 105

<=> 38x = 0

<=> x = 0

Vậy x = 0

c) 32 (x - 11) + 4x = 152

<=> 32x - 352 + 4x = 152

<=> 36x = 504

<=> x = 14

Vậy x = 14

d) 51 ( 3x + 5) - 406 = 22 (3x +50)

<=> 153x + 255 - 406 = 66x + 1100

<=> 87x = 1251

<=> x = \(\frac{1251}{87}\)

Vậy x = \(\frac{1251}{87}\)

Học tốt nhá :))

Bài làm :

\(a,17x+33x=100\)

\(\left(17+33\right)x=100\)

\(50x=100\)

\(x=2\)

\(b,23x+15\left(x+7\right)=105\)

\(23x+15x+105=105\)

\(38x=0\)

\(x=0\)

\(c,32\left(x-11\right)+4x=152\)

\(32x-352+4x=152\)

\(36x=504\)

\(x=14\)

\(d,51\left(3x+5\right)-406=22\left(3x+50\right)\)

\(153x+255-406=66x+1100\)

\(153x-66x=1100+406-255\)

\(87x=1251\)

\(x=\frac{1251}{87}\)

Học tốt

Bài 1.

x = 14

=> 13 = x - 1 ; 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1

Thế vào N(x) ta được :

x⁵ - ( x + 1 )x⁴ + ( x + 2 )x³ - ( 2x + 1 )x² + ( x - 1 )x

= x⁵ - x⁵ - x⁴ + x⁴ + 2x³ - 2x³ - x² + x² - x

= -x = -14

Bài 2.

a) ( 1 - x - 2x³ + 3x² )( 1 - x + 2x³ - 3x² )

= [ ( 1 - x ) - ( 2x³ - 3x² ) ][ ( 1 - x ) + ( 2x³ - 3x² ) ]

= ( 1 - x )² - ( 2x³ - 3x² )²

= 1 - 2x + x² - [ ( 2x³ )² - 2.2x³.3x² + ( 3x² )² ]

= x² - 2x + 1 - ( 4x⁶ - 12x⁵ + 9x⁴ )

= x² - 2x + 1 - 4x⁶ + 12x⁵ - 9x⁴

= -4x⁶ + 12x⁵ - 9x⁴ + x² - 2x + 1

b) ( x - y + z )² + ( z - y )² + 2( x - y + z )( y - z )

= ( x - y + z )² + ( z - y )² - 2( x - y + z )( z - y )

= [ ( x - y + z ) - ( z - y ) ]²

= ( x - y + z - z + y )²

= x²

\(P=\left(m^2-5m+\frac{25}{4}\right)-\frac{13}{4}=\left(m-\frac{5}{2}\right)^2-\frac{13}{4}\)

Vì \(m\ge3 \Rightarrow m-\frac{5}{2}\ge\frac{1}{2} \Rightarrow\left(m-\frac{5}{2}\right)^2\ge\frac{1}{4} \Rightarrow\left(m-\frac{5}{2}\right)^2-\frac{13}{4}\ge\frac{1}{4}-\frac{13}{4}\)

\(\Rightarrow P\ge-3\)

\(MinP=-3\Leftrightarrow m=3\)

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mạch điện ntn bạn ?

\(I_{A_2}=1,5\left(A\right)?\)

Ta có: \(\frac{R_1}{R_2}=\frac{I_{A_2}}{I_{A_1}}\)

\(\Leftrightarrow\frac{20}{R_2}=\frac{1,5}{0,5}\)

\(\Rightarrow R_2=\frac{20}{\frac{1,5}{0,5}}=\frac{20}{3}\left(\Omega\right)\)

Vậy \(R_2=\frac{20}{3}\Omega\)

a. ( x + 1 ) ( x + 2 ) ( x - 3 )

= ( x² + 3x + 2 ) ( x - 3 )

= x³ + 3x² + 2x - 3x² - 9x - 6

= x³ - 7x - 6

b. ( 2x - 1 ) ( x + 2 ) ( x + 3 )

= ( 2x² + 3x - 2 ) ( x + 3 )

= 2x³ + 3x² - 2x + 6x² + 9x - 6

= 2x³ + 9x² + 7x - 6

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

= ( x² + 3x + 2 )( x - 3 )

= x³ - 3x² + 3x² - 9x + 2x - 6

= x³ - 7x - 6

b) ( 2x - 1 )( x + 2 )( x + 3 )

= ( 2x² + 3x - 2 )( x + 3 )

= 2x³ + 6x² + 3x² + 9x - 2x - 6

= 2x³ + 9x² + 7x - 6

\(BPT\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(3x^2+2x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(x+1\right)\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(x+1\right)\text{[}2+\sqrt{x^2-2x+5}+\frac{2x\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\text{]}\le0\)

\(\Leftrightarrow\left(x+1\right)\left(4\sqrt{x^2+1}+2\sqrt{x^2-2x+5}+2\sqrt{\left(x^2+1\right)\left(x^2-2x+5\right)}+7x^2-4x+5\right)\)\(\le0\Leftrightarrow x+1\le0\Leftrightarrow x\le-1\)

đk: \(x\ne0\)

Ta có:

\(D=x+\sqrt{x^2+\frac{1}{x^2}+2}\)

\(D=x+\sqrt{\left(x+\frac{1}{x}\right)^2}\)

\(D=x+\left|x+\frac{1}{x}\right|\)

\(D=x-x-\frac{1}{x}\) \(\left(x< 0\right)\)

\(D=-\frac{1}{x}\)