a.x + 2/5 = 1 - 1/6

x + 2/5 = 5/6

x= 13/30

b.3/4 - 2/5 + 5/12

=7/20 + 5/12

=23/30

a)\(x=\left(1-\dfrac{1}{6}\right)-\dfrac{2}{5}=\dfrac{5}{6}-\dfrac{2}{5}=\dfrac{13}{30}\)

b)\(=\dfrac{7}{20}+\dfrac{5}{12}=\dfrac{23}{30}\)

a.x + 2/5 = 1 - 1/6

x + 2/5 = 5/6

x= 13/30

b.3/4 - 2/5 + 5/12

=7/20 + 5/12

=23/30

\(\Leftrightarrow4x-8+7⋮x-2\)

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

hay \(x\in\left\{3;1;9;-5\right\}\)

⇔4x−8+7⋮x−2⇔4x−8+7⋮x−2

hay 

c: =>5x=40

hay x=8

⇔5x=6²+4

⇔5x=36+4

⇔5x=40

⇔x=8

Đây bạn!!!
c) 9x – 4x = 6¹⁷ : 6¹⁵ + 48 : 12
=> 5x = 6²+4
=> 5x = 36+4
=> 5x = 40
=> x = 8
 

a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)

\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)

\(\Leftrightarrow-36x=72\)

hay x=-2

b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)

\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)

\(\Leftrightarrow4x=96\)

hay x=24

c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)

\(\Leftrightarrow x^2+3x-4-x^2+x=308\)

\(\Leftrightarrow4x=312\)

hay x=78

d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)

\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)

\(\Leftrightarrow-32x=-32\)

hay x=1

Đề là gì v cậu?

\(a\left(\sqrt{2}-1\right)+b\left(\sqrt{2}+1\right)=12\\ \Leftrightarrow a\sqrt{2}-a+b\sqrt{2}+b=12\)

Đề như vậy á cậu?

Áp dụng dãy tỉ số bằng nhau:

b.

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)

d.

\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)

(2x+1).(y2-5)=12=1.12=12.1=6.2=2.6=3.4=4.3=...(cả số âm)

Rồi bạn lập bảng

VD:

`(2x+1)(y^2-5)=12=1.12=(-1).(-12)=2.6=(-2).(-6)=3.4=(-3).(-4)`

  Vì `x;y` là số tự nhiên `=>x=1;y=3`

2. Ta có: A = x² - 6x + 5 = (x² - 6x + 9) - 4 = (x - 3)² - 4 

Ta luôn có: (x - 3)² \(\ge\)0 \(\forall\)x

=> (x - 3)² - 4 \(\ge\)-4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3

Vậy MinA = -4 tại  x = 3

Ta có: B = 4x² - 8x + 7 = 4(x² - 2x + 1) + 3 = 4(x - 1)² + 3

Ta luôn có: 4(x - 1)² \(\ge\)0 \(\forall\)x

=> 4(x - 1)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1

vậy MinB = 3 tại x = 1

Ta có: C = 2x² + 4x - 6 = 2(x² + 2x + 1) - 8 = 2(x + 1)² - 8

Ta luôn có: 2(x + 1)² \(\ge\)0 \(\forall\)x

=> 2(x + 1)² - 8 \(\ge\)-8 \(\forall\)x

Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1

Vậy MinC = -8 tại x = -1

1/

\(A=x^2-6x+5\)

\(A=x^2-2\cdot3x+3^2-3^2+5\)

\(A=\left(x-3\right)^2-3^2+5\)

\(A=\left(x-3\right)^2-9+5\)

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

mà \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-4\ge-4\)

\(\Rightarrow GTNNA\left(x^2-6x+5\right)=-4\)

với \(\left(x-3\right)^2=0;x=3\)

\(B=4x^2-8x+7\)

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

\(B=4\left(x^2-2\cdot1x+1-1+\frac{7}{4}\right)\)

\(B=4\left(x-1\right)^2+3\)

\(\left(x-1\right)^2\ge0\Rightarrow4\left(x^2-1\right)^2+3\ge3\)

\(\Rightarrow GTNNB=3\)

với \(\left(x-1\right)^2=0;x=1\)

\(C=2x^2+4x-6\)

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

\(C=2\left(x^2+2\cdot1x+1-1-3\right)\)

\(C=\left(x+1\right)^2-8\)

có\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2-8\ge-8\)

\(\Rightarrow GTNNC=-8\)

với \(\left(x+1\right)^2=0;x=-1\)

3

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

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

\(A=-\left(x^2-2\cdot1x+1-1+3\right)\)

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

có \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)

\(\Rightarrow GTLNA=-2\) với \(\left(x-1\right)^2=0;x=1\)

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

\(B=-9\left(x^2-\frac{2}{3}x+\frac{4}{9}\right)\)

\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{1}{9}+\frac{4}{9}\right)\)

\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{3}{9}\right)\)

\(B=-9\left(x-\frac{1}{3}\right)^2+3\)

có \(\left(x-\frac{1}{3}\right)^2\ge0\Rightarrow-9\left(x-\frac{1}{3}\right)^2+3\ge3\)

\(\Rightarrow GTLNB=3\)với \(\left(x-\frac{1}{3}^2=0;\right)x=\frac{1}{3}\)