`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

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

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

này sao lại mũ 29 rồi lại 20

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

câu 1:

đường thẳng (d) song song với đường thẳng y=3x+1 khi a=3

vậy hệ số góc của đường thẳng (d) song song với đường thẳng y=3x+1 là 3

câu 2:

vì góc tạo bởi đường thẳng (d):y=ax+b(a≠0) với trục Ox là 30^o nên

\(a=\tan30^o=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

vậy hệ số góc của đường thẳng (d) tạo với trục Ox là\(\dfrac{\sqrt{3}}{3}\)

Bài 1:

a) A= x² + 4x + 5

=x²+4x+4+1

=(x+2)²+1\(\ge\)0+1=1

Dấu = khi x+2=0 <=>x=-2

Vậy Amin=1 khi x=-2

b) B= ( x+3 ) ( x-11 ) + 2016

=x²-8x-33+2016

=x²-8x+16+1967

=(x-4)²+1967\(\ge\)0+1967=1967

Dấu = khi x-4=0 <=>x=4

Vậy Bmin=1967 <=>x=4

Bài 2:

a) D= 5 - 8x - x² 

=-(x²+8x-5)

=21-x²+8x+16

=21-x²+4x+4x+16

=21-x(x+4)+4(x+4)

=21-(x+4)(x+4)

=21-(x+4)²\(\le\)0+21=21

Dấu = khi x+4=0 <=>x=-4

b)đề sai à

ài 1:

a) A= x² + 4x + 5

=x²+4x+4+1

=(x+2)²+1$\ge$≥0+1=1

Dấu = khi x+2=0 <=>x=-2

Vậy Amin=1 khi x=-2

b) B= ( x+3 ) ( x-11 ) + 2016

=x²-8x-33+2016

=x²-8x+16+1967

=(x-4)²+1967$\ge$≥0+1967=1967

Dấu = khi x-4=0 <=>x=4

Vậy Bmin=1967 <=>x=4

Bài 2:

a) D= 5 - 8x - x² 

=-(x²+8x-5)

=21-x²+8x+16

=21-x²+4x+4x+16

=21-x(x+4)+4(x+4)

=21-(x+4)(x+4)

=21-(x+4)²$\le$≤0+21=21

Dấu = khi x+4=0 <=>x=-4

b)đề sai à

a,

(x²-x+1)(x+1)-x³+3x=15

x³-x²+x+x²-x+1-x³+3x=15

x³-x³-x²+x²+x-x+3x+1=15

3x+1=15

3x=15-1

3x=14

x=14/3

b,

(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)

x²-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)

Tới đây hết biết , đề có gì sai sai sao ý !

c,

(x²-5)(x+2)+5x=2x²+17

x³+2x²-5x-10+5x=2x²+17

x³+2x²-5x+5x-10=2x²+17

x³+2x²-10=2x²+17

x³-10=17

x³=17+10

x³=27

\(\Rightarrow x=3\)(Vì : 3³=27)

_k_ nhé bn

Nhân ra thôi bạn, có hằng đẳng thức gì đâu !

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

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

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

\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)

b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)

\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)

\(\Leftrightarrow x^2+4x-6=4x+3\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)

c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)

\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)

\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)