Mình trả lời hai câu đầu trước nha!

1)

x= -6+2

x= -4

2)

=-5 x = 13 + (-3)

= -5 .x =10

x = 10 ÷ -5

x = -2

a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)

b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)

c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)

\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)

d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)

\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)

a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)

<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)

<=> \(\sqrt{x}+8=28\)

<=> \(\sqrt{x}=28-8\)

<=> \(\sqrt{x}=20\)

<=> \(\left(\sqrt{x}\right)^2=20^2\)

<=> x = 400

=> x = 400

b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)

<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)

<=> \(3\sqrt{x}+5=\sqrt{x}+12\)

<=> \(3\sqrt{x}=\sqrt{x}+12-5\)

<=> \(3\sqrt{x}=\sqrt{x}+7\)

<=> \(3\sqrt{x}-\sqrt{x}=7\)

<=> \(2\sqrt{x}=7\)

<=> \(\sqrt{x}=\frac{7}{2}\)

<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)

<=> \(x=\frac{49}{4}\)

=> \(x=\frac{49}{4}\)

c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)

<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)

<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)

<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)

<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)

<=> \(8\sqrt{x}=6\sqrt{x}+4\)

<=> \(8\sqrt{x}-6\sqrt{x}=4\)

<=> \(2\sqrt{x}=4\)

<=> \(\sqrt{x}=2\)

<=> \(\left(\sqrt{x}\right)^2=2^2\)

<=> x = 4

=> x = 4

d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)

<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)

<=>\(2\sqrt{3x}=6\sqrt{3x}\)

<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)

<=>\(-4\sqrt{3x}=0\)

<=> \(\sqrt{3x}=0\)

<=> \(\left(\sqrt{3x}\right)^2=0^2\)

<=> 3x = 0

<=> x = 0

=> x = 0

à, phần a ra x = 400. Nhầm

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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a) 71 + \(\frac{26-3x}{5}=75\)

↔ \(\frac{26-3x}{5}=4\)

↔\(26-3x=20\)

↔x=2

b) \(x-17=\left(-8\right)-17\)

↔ \(x-17=-25\)

↔\(x=-8\)

c)\(\left(5x-2\right).6^3=3.6^5\)

↔\(\left(5x-2\right)=3.6^2\)

↔\(5x=110\)

↔\(x=22\)

d)\(\left(3x-2^4\right).7^3=2.7^4\)

↔\(3x-16=14\)

↔x=10

tick nha~~

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

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

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

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

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a: \(\Leftrightarrow\left(-x+3\right)\left(x+6\right)=18\)

\(\Leftrightarrow-x^2-6x+3x+18-18=0\)

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

=>x=0 hoặc x=-3

b: \(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2x-\dfrac{4}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{\sqrt{21}}{3}-1;\dfrac{-\sqrt{21}}{3}-1\right\}\)

c: =>x(3x-5)=0

=>x=0 hoặc x=5/3

d: =>(x-2)(x+2)=0

=>x=2 hoặc x=-2

Ta có : x(x - 3) - 2x + 6 = 0

<=> x(x - 3) - (2x - 6) = 0

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)

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