a)Ta có:

\(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

b) Ta có:

\(x\left(x-4\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

a) 7(x - 5) + 2 = 51

\(\Leftrightarrow\) 7(x - 5) = 51 - 2

\(\Leftrightarrow\) 7(x - 5) = 49

\(\Leftrightarrow\) x - 5 = 49 : 7

\(\Leftrightarrow\) x - 5 = 7

\(\Leftrightarrow\) x = 7 + 5

\(\Leftrightarrow\) x = 12.

Vậy x = 12.

k) 2412 : (3x + 147) = |-38| + (-26)

\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)

\(\Leftrightarrow\) 2412 : (3x + 147)= 12

\(\Leftrightarrow\) 3x + 147 = 2412 : 12

\(\Leftrightarrow\) 3x + 147 = 201

\(\Leftrightarrow\) 3x = 201 - 147

\(\Leftrightarrow\) 3x = 54

\(\Leftrightarrow\) x = 54 : 3

\(\Leftrightarrow\) x = 18.

Vậy x = 18.

I) 4824 : (4x + 137) = |-59| + (-35)

\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)

\(\Leftrightarrow\) 4824 :(4x + 137) = 24

\(\Leftrightarrow\) 4x + 137 = 4824 : 24

\(\Leftrightarrow\) 4x + 137 = 201

\(\Leftrightarrow\) 4x = 201 - 137

\(\Leftrightarrow\) 4x = 64

\(\Leftrightarrow\) x = 64 : 4

\(\Leftrightarrow\) x = 16.

Vậy x = 16.

c) |-123| - 5(x - 3) = (-28) + 66

\(\Leftrightarrow\) 123 - 5(x - 3) = 38

\(\Leftrightarrow\) 5(x - 3) = 123 - 38

\(\Leftrightarrow\) 5(x - 3) = 85

\(\Leftrightarrow\) x - 3 = 85 : 5

\(\Leftrightarrow\) x - 3 = 17

\(\Leftrightarrow\) x = 17 + 3

\(\Leftrightarrow\) x = 20.

Vậy x = 20.

m) 7x - 4 . 6 = 2058

\(\Leftrightarrow\) 7x - 24 = 2058

\(\Leftrightarrow\) 7x = 2058 + 24

\(\Leftrightarrow\) 7x = 2082

\(\Leftrightarrow\) x = 2082 : 7

\(\Leftrightarrow\) x = \(\frac{2058}{7}\).

Vậy x = \(\frac{2058}{7}\).

Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.

Chúc bạn học tốt !!!

---------------NHANH NHA MK ĐANG CẦN GẤP--------------

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g: =>3x-10=200

hay x=70

c) (7x + x – 28) :100 – 38 = 9

<=> 8x - 28 = 4700

<=> x = 591

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)

a: =>x/27+1=-2/3

=>x/27=-5/3

=>x=-45

b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)

=>x=221/50

c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)

=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)

=>x=-37/24

e: =>-3/7x=84/45

=>x=-196/45

f: =>11/10x=-2/3

=>x=-20/33

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

Sao lm mỗi câu a, c thế

Câu c viết dấu hoặc đi

1.

4x(7x-5)-7x(4x-2)=12

28x²-20x-28x²+14x=12

28x²-28x²-20x+14x=12

-6x=12

x=12:-6

x=-2

2.

Từ mình làm tiếp

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)