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\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(=4x^2+1-4x+\left(x^2+9+6x\right)-5\left(x^2-7^2\right)=0\)

\(=4x^2+1-4x+x^2+9+6x-5x^2+245=0\)

\(=\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+245\right)=0\)

\(=2x+255=0\)

\(\Rightarrow2x=-255\)

\(x=-127,5\)

(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0

=>4x²-4x+1+x²+6x+9+245-5x²=0

=>(4x²+x²-5x²)+(6x-4x)+(1+9+245)=0

=>2x+255=0

=>2x=-255 <=>x=-255/2

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(=4x^2+1-4x+\left(x^2+9+6x\right)-5\left(x^2-7^2\right)=0\)

\(=4x^2+1-4x+x^2+9+6x-5x^2+245=0\)

\(=\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+245\right)=0\)

\(=2x+255=0\)

\(\Rightarrow2x=-255\)

\(x=-127,5\)

a)

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9=3^2\)

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

\(\Rightarrow\left(5x+1\right)^2=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

b/ x²-2x=24

=> x²-2x-24=0

=> (x-6)(x+4)=0

=>x=6 hoặc x =-4

a/ (x-3)² - 4 = 0

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

=> (x-5)(x-1)=0

=> x = 5 hoặc x=1

ý c đâu ạ

Bạn đã học hằng đẳng thức chưa để mk giải 

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

1) \(x^2-9x=0\Rightarrow x\left(x-9\right)=0\Rightarrow x=0;9\)

2) \(x\left(x-4\right)-x^2=7\Rightarrow-4x=7\Rightarrow x=-\dfrac{7}{4}\)

3) \(3x+2\left(x-5\right)=5\Rightarrow5x-10=5\Rightarrow5x=15\Rightarrow x=3\)

4) \(25x^2-1=0\Rightarrow x^2=\dfrac{1}{25}\Rightarrow x=\pm\dfrac{1}{5}\)

5) \(3x\left(x-2\right)-5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-5\right)=0\Rightarrow x=2;\dfrac{5}{3}\)

6) \(3x\left(x-7\right)+4\left(x-7\right)\Rightarrow\left(3x+4\right)\left(x-7\right)=0\Rightarrow x=-\dfrac{4}{3};7\)

7) \(4x^2-9=0\Rightarrow x^2=\dfrac{9}{4}\Rightarrow x=\pm\dfrac{3}{2}\)

8) \(10x\left(x-4\right)+2x-8=0\Rightarrow2\left(x-4\right)\left(5x+1\right)=0\Rightarrow x=4;-\dfrac{1}{5}\)

9) \(x\left(2x-5\right)-2x^2=0\Rightarrow x\left(2x-5-2x=0\right)\Rightarrow x=0\)

10) \(2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\Rightarrow x=0;2\)

11) \(2x\left(3-4x\right)+3\left(4x-3\right)=0\Rightarrow2x\left(4x-3\right)-3\left(4x-3\right)=0\Rightarrow\left(4x-3\right)\left(2x-3\right)=0\Rightarrow x=\dfrac{3}{4};\dfrac{3}{2}\)

12) \(2x\left(x-5\right)-2x^2=3\Rightarrow-10x=3\Rightarrow x=-\dfrac{3}{10}\)

1) (x - 35) - 120 = 0

x - 35 = 120

x = 120 + 35

x = 155

2) 310 - (118 - x) = 217

118 - x = 310 - 217

118 - x = 93

x = 118 - 93

x = 25

3) 156 - (x + 61) = 82

x + 61 = 156 - 82

x + 61 = 74

x = 74 - 61

x = 13

4) 814 - (x - 305) = 712

x - 305 = 814 - 712

x - 305 = 102

x = 102 + 305 = 407

5) 100 - 7 - (x - 5) = 58

x - 5 = 93 - 58

x - 5 = 35

x = 35 + 5 = 40

6) 12(x - 1) : 3 = 4³ + 2³

4(x - 1) = 72

x - 1 = 18

x = 18 + 1 = 19

7) 24 + 5x = 7⁵ : 7³

24 + 5x = 49

5x = 25

x = 25 : 5 = 5

8) 5(x - 1) : 3 = 4³ + 2³

\(\dfrac{5}{3}\left(x-1\right)=72\)

x - 1 = \(\dfrac{216}{5}\)

x = 221/5

9) 5(x - 4)² - 7 = 13

5(x - 4)² = 20

(x - 4)² = 4

\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\)   \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

10) (x + 1) + (x + 2) + ... + (x + 30) = 795

=> (x + x + x + ... + x) + (1 + 2 + 3 +...+ 30) = 795 (1)

Đặt A = 1 + 2 + 3 +...+ 30

Số số hạng trong A là: (30 - 1) : 1 + 1 = 30 (số)

Tổng A bằng : (30 + 1).30 : 2 =465

Thay A = 465 vào (1) , ta được:

30x + 465 = 795

=> 30x =330

=> x =11

1: =>x-35=120

=>x=120+35=155

2: =>118-x=310-217=93

=>x=118-93=25

3: =>x+61=156-82=74

=>x=74-61=13

4: =>x-305=814-712=102

=>x=102+305=407

5: =>93-(x-5)=58

=>x-5=35

=>x=40

6: =>4(x-1)=64+8=72

=>x-1=18

=>x=19

7: =>5x+24=49

=>5x=25

=>x=5

8: =>5(x-1):3=4^3+2^3=64+8=72

=>5(x-1)=216

=>x-1=216/5

=>x=221/5

a) x = 4

b) x = 2 ; x = -4

c) x = 2 ; x = -15

d) x = 7 ; x = -19

e) x = -4 ; -3 ; -2 ; -1 ; 0

g) x = -1 ; - 2 ; 1 ; 2 ; 3 ; 4 ; ...

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a) -12x + 60 + 21 - 7x = 5

-12x - 7x + 60+21 = 5

-19x + 81 = 5

81-5 = 19x

19x = 76

x= 4

nhầm đề bài ak