b: \(\Leftrightarrow8x=41\)

hay x=41/8

a) (x - 2)² - (x + 3) - 4(x + 1) = 5

=> (x² - 4x + 4) - (x + 3) - 4x - 4 = 5

=> x² - 4x + 4 - x - 3 - 4x - 4 = 5

=> x² + (-4x - x - 4x) + (4 - 3 - 4) = 5

=> x² - 9x - 3 = 5

=> x² - 9x - 3 - 5 = 0

=> x² - 9x - 8 = 0

=> [x² - 2.x.9/2 + (9/2)² ] - 113/4 = 0

=> (x - 9/2)² - 113/4 = 0

=> (x - 9/2)² - \(\left(\sqrt{\frac{113}{4}}\right)^2\)= 0

=> \(\left(x-\frac{9}{2}-\sqrt{\frac{113}{4}}\right)\left(x+\frac{9}{2}+\sqrt{\frac{113}{4}}\right)=0\)

=> \(\hept{\begin{cases}x-\frac{9}{2}-\sqrt{\frac{113}{4}}=0\\x+\frac{9}{2}+\sqrt{\frac{113}{4}}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9+\sqrt{133}}{2}\\x=-\frac{9+\sqrt{133}}{2}\end{cases}}\)

Nếu không muốn nghiệm xấu như thế này thì bạn để vô nghiệm

b) (2x - 3)(2x + 3) - (x - 1)² - 3x(x - 5) = -44

=> (2x)² - 9 - (x² - 2x + 1) - 3x² + 15x = -44

=> 4x² - 9 - x² + 2x - 1 - 3x² + 15x = -44

=> (4x² - x² - 3x²) + (2x + 15x) + (-9 - 1) = -44

=> 17x  - 10 = -44

=> 17x = -34

=> x = -2

Vậy x = -2

a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)

\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1\right\}\)

b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)

\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)

\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)

\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)

\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)

\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)

hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)

Bài 1:

x/-3=9/4

nên x=-9/4*3=-27/4

2x+y=-4

=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a, \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=16\)

\(\Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=16\)

\(\Leftrightarrow8x-4=16\Leftrightarrow x=\frac{20}{8}=\frac{5}{2}\)

b, \(3x\left(2x-4\right)-2x\left(3x+5\right)=44\)

\(\Leftrightarrow6x^2-12x-\left(6x^2-10x\right)=44\Leftrightarrow-2x=44\Leftrightarrow x=-22\)

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

\(\Leftrightarrow\left(10x-22\right)\left(4x-5\right)=4\left(3x-4\right)\Leftrightarrow40x^2-50x-88x+110=0\)

\(\Leftrightarrow40x^2-138x+110=0\)( vô nghiệm ) 

Cái dề, bài này giúp á, tính ra là ok mà

(13/10x - 15).(-5/14) = 3

13/10x - 15 = 3 : (-5/14)

13/10x - 15 = 3 x (-14/5)

13/10x - 15 = -42/5

=> 13/10x = -42/5 + 15

=> 13/10x = 33/5

=> x = 33/5 : 13/10

=> x = 33/5 x 10/13

=> x = 66/13

bn tự lm típ, tính ra là ok mà

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

7x-2x=6^17:6^15+44:11

=>2x=6^(17-15)+4

=>2x=6^2+4

=>2x=36+4

=>2x=40

=>x=40:2

=>x=20

5x=6^2+4

5x=40

x=40:5=8

Vậy x=8

7x - 2x = 6^17 : 6^15 + 44 : 11

5x         = 6^2 + 4

5x          = 36+4

5x            = 40

x              = 8

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

\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)

\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)

\(\Leftrightarrow-14x-9=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

Vậy....

b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow-10+17x=-44\)

\(\Leftrightarrow17x=-34\)

\(\Leftrightarrow x=-2\)

Vậy....

c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy....

d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)

\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)

\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)

\(\Leftrightarrow14x-3=7\)

\(\Leftrightarrow14x=10\)

\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)

Vậy...

câu hỏi đây

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