(x-42) - 17 = 127

=> x - 42 = 127 + 17 = 144

=> x = 144 + 42 = 186

23(x+1) = 69

=> x + 1 = 69 : 23 = 3

x = 3 - 1 = 2

2x + 5 = 120 : 2 = 60

=> 2x = 60 - 5 = 55

x = 55 : 2 = 27,5

5x - 2 = 613

=> 5x = 613 + 2 = 615

x = 615 : 5 = 123

a)(x-42)-17=127

(x-42)=127+17

(x-42)=144

x=144+42

x=186

b)23(x+1)=69

(x+1)=69:23

(x+1)=3

x=3-1

x=2

c)2.x+5=120:2

2.x+5=60

2.x=60-5

2.x=55

x=55:2

x=27,5

d)5.x-2=613

5.x=613+2

5.x=615

x=615:5

x=123

(x-42)-17=127(tim x)xx

23(x+1)=69(timx)xx

2x+5=120:2(Timx)xx

5x-2=613(timx)xx

\((x-2)(x^2+2x+5+2x+4-5)=0<=> (x-2)(x^2+4x+4)=0 <=> (x-2)(x+2)^2 = 0 <=> x = 2 ; x = -2\)

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

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+5+2x+4-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+4\right)=0\)

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

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

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

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

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

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 \(\left(x;y\right)=\left(2;-3\right)\)

Bài 1 :

\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)

\(\Leftrightarrow x=-23\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)

(3x+2).(x+1)=3x.(5+x)                                                                               

\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)                               

\(\Rightarrow3x^2+5x+2=15x+3x^2\)

\(\Rightarrow5x-15x+2=3x^2-3x^2\)

\(\Rightarrow-10x+2=0\)

\(-10x=-2\)

    \(x=\frac{1}{5}\)

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

=(x+1+x-1)(x+1-x+1)-3x^2-3

=2x^2-3x^2-3

=-x^2-3

mở dấu trị tuyệt đối ra rồi tính như bình thường

\(\Leftrightarrow x^2+8x+16+x^2-2x+1-2\left(x^2-25\right)=21\\ \Leftrightarrow2x^2+6x+17-2x^2+50=21\\ \Leftrightarrow6x=-46\Leftrightarrow x=-\dfrac{23}{3}\)

a) (5 + x)(x - 5) - x(x + 5) = 10

x² - 25 - x² - 5x = 10

-5x = 10 + 25

-5x = 35

x = 35 : (-5)

x = -7

b) x.(2x + 3) - 2(x² + x) = 2

2x² + 3x - 2x² - 2x = 2

x = 2

a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)

=>\(x^2-25-x^2-5x=10\)

=>-5x-25=10

=>-5x=35

=>x=-7

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

=>\(2x^2+3x-2x^2-2x=2\)

=>x=2