Mấy bài này căng vậy?

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

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

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

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

~Hok tốt~

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

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x = 2/7

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

a) Ta có: \(A=x^2-6x+11\)

\(=x^2-6x+9+2\)

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

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

Ta có: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi

\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy: GTNN của đa thức \(A=x^2-6x+11\) là 2 khi x=3

b) Ta có: \(B=x^2-4x+3\)

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

\(=\left(x^2-4x+4\right)-1\)

\(=\left(x-2\right)^2-1\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2-1\ge-1\forall x\)

Dấu '=' xảy ra khi

\(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy: GTNN của đa thức \(B=x^2-4x+3\) là -1 khi x=2

c) Ta có: \(C=x^2+5x\)

\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)

\(=\left(x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)

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

Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)

Dấu '=' xảy ra khi

\(\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)

Vậy: GTNN của đa thức \(C=x^2+5x\) là \(\frac{-25}{4}\) khi \(x=\frac{-5}{2}\)

d) Ta có: \(D=x^2+x+1\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu '=' xảy ra khi

\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)

Vậy: GTNN của đa thức \(D=x^2+x+1\) là \(\frac{3}{4}\) khi \(x=\frac{-1}{2}\)

e) Ta có: \(E=4x^2+4x-2\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1-3\)

\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1\right]-3\)

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

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+1\right)^2-3\ge-3\forall x\)

Dấu '='xảy ra khi

\(\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)

Vậy: GTNN của đa thức \(E=4x^2+4x-2\) là -3 khi \(x=\frac{-1}{2}\)

g) Ta có: \(G=x^2-7x\)

\(=x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{14}-\frac{49}{14}\)

\(=\left(x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{4}\right)-\frac{49}{4}\)

\(=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\)

Ta có: \(\left(x-\frac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\ge\frac{-49}{4}\forall x\)

Dấu '=' xảy ra khi

\(\left(x-\frac{7}{2}\right)^2=0\Leftrightarrow x-\frac{7}{2}=0\Leftrightarrow x=\frac{7}{2}\)

Vậy: GTNN của đa thức \(G=x^2-7x\) là \(\frac{-49}{4}\) khi \(x=\frac{7}{2}\)

\(A=x^2-6x+11\)

\(A=x^2-2.x.3+3^2-3^2+11\)

\(A=\left(x^2-6x+3^2\right)-3^2+11\)

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

Vì \(\left(x-3\right)^2\ge0\forall x\)

=>\(\left(x-3\right)^2\ge0\ge2\forall x\)

Min A = 2 khi \(\left(x-3\right)^2=0\)

=> \(x-3=0hayx=3\)

Vậy Min A = 2 khi x = 3

\(B=x^2-4x+3\)

\(B=x^2-2.x.2+2^2-2^2+3\)

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

\(B=\left(x-2\right)^2-1\)

=> \(\left(x-2\right)^2-1\ge0\forall x\)

MIn B = -1 khi \(\left(x-2\right)^2=0\)

=>\(\left(x-2\right)=0hayx=2\)

Vậy Min B = -1 khi x= 2

a) (x-3).(y+5) = 11 = 1.11 = (-1).(-11)

TH1: x - 3 = 1 => x = 4

y + 5 = 11 => y = 6

TH2: x - 3 = 11 => x = 14

y+5=1 => y = -4

...

bn tự lm típ nhé!

b) |x-1| +|3+y| = 0

=> |x-1| = 0 =>x-1 = 0 => x = 1

|3+y| = 0 => 3+y = 0=> y = - 3

c) ta có: 4x+3 chia hết cho x - 1

=> 4x -4+7 chia hế cho x - 1

4.(x-1) + 7 chia hết cho x - 1

mà 4.(x-1) chia hết cho x - 1

=> 7 chia hết cho x - 1

=> x - 1 thuộc Ư(7)={1;-1;7;-7}

...

rùi bn lập bảng xét giá trị hộ mk nha!!
 

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

\(-\left(-a+b+c\right)+\left(b-c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(a-b-c+b-c-1=b-c+6-7+a-b+c\)

\(a-2c-1=a-1\)

\(-2c\ne0\)hay đẳng thức ko xảy ra 

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

a: \(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

=>\(-4x^2+20x-16x+4x^2=-3\)

=>4x=-3

=>\(x=-\dfrac{3}{4}\)

b: \(-7\left(x+9\right)-3\left(5-x\right)=2\)

=>\(-7x-63-15+3x=2\)

=>\(-4x-78=2\)

=>\(-4x=78+2=80\)

=>\(x=\dfrac{80}{-4}=-20\)