\(\frac{2}{x}=\frac{y}{15}=\frac{\left(-25\right)}{75}\)

\(\Rightarrow\frac{y}{15}=\frac{\left(-1\right)}{3}\)

\(\frac{y}{15}=\frac{\left(-5\right)}{15}\)

\(\Rightarrow15y=15\times\left(-5\right)\)

\(15y=\left(-75\right)\)

\(y=\left(-75\right)\div15\)

\(y=\left(-5\right)\)

\(\Rightarrow\frac{2}{x}=\frac{\left(-5\right)}{15}\)

\(\frac{2}{x}=\frac{\left(-1\right)}{3}\)

\(\Rightarrow-x=3\times2\)

\(-x=6\)

\(x=\left(-6\right)\)

\(V\text{ậy}x=\left(-6\right);y=\left(-5\right)\)

Số hạng tử X:

(100 - 0): 2 + 1 = 51 (hạng tử)

Tổng hạng tử số:

(100+2) x (50:2)= 2550

Tổng của 51 hạng tử X:

2601 - 2550=51

X có giá trị bằng:

51:51=1

Vậy x=1

` @ L I N H `

Số hạng tử X:

(100 - 0): 2 + 1 = 51 (hạng tử)

Tổng hạng tử số:

(100+2) x (50:2)= 2550

Tổng của 51 hạng tử X:

2601 - 2550=51

X có giá trị bằng:

51:51=1

Vậy x=1

0o0 bạn học giỏi ghê !  thank you nha ! ❤

a) Ta có: \(\left|-5\right|+\left|x-1\right|=\left|7\right|\)

\(\Leftrightarrow\left|x-1\right|+5=7\)

\(\Leftrightarrow\left|x-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-1\right\}\)

b) Ta có: \(2\cdot\left|2x-4\right|-\left|-4\right|=\left|-50\right|\)

\(\Leftrightarrow4\cdot\left|x-2\right|-4=50\)

\(\Leftrightarrow4\cdot\left|x-2\right|=54\)

\(\Leftrightarrow\left|x-2\right|=\dfrac{27}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=\dfrac{27}{2}\\x-2=-\dfrac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{31}{2}\left(loại\right)\\x=-\dfrac{23}{2}\left(loại\right)\end{matrix}\right.\)

Vậy: \(x\in\varnothing\)

a, | -5 | + | x-1 | = | 7 |

         5 + | x - 1 | = 7 

               | x - 1 | = 2

TH1  x -1 = 2

        x = 3

TH2 x -1 = -2

        x= -1

Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)

\(\left(y+0.4\right)^{100}\ge0\forall y\)

\(\left(z-3\right)^{678}\ge0\forall z\)

Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)

Câu 1 : A = 1+1+1+...+1 ( 50 số 1 ) = 50

1. 

A = x² + x⁴ + x⁶ + ... + x¹⁰⁰ ( 50 số hạng )

A = ( -1 )² + ( -1 )⁴ + ( -1 )⁶ + ... + ( -1 )¹⁰⁰

A = 1 + 1 + 1 + ... + 1

A = 50

2. 

| x - 1/3 | + 4/5 = | (-3,2) + 2/50 |

| x - 1/3 | + 4/5 = 3,16

| x - 1/3 | = 2,36

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2,36\\x-\frac{1}{3}=-2,36\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{202}{75}\\x=\frac{-152}{75}\end{cases}}\)

Quên mất, mọi người giải có công thức tính hộ mình nha!

\(\frac{75}{100}+\frac{3}{4}x29+75\%x30+0,75x40\)

\(=\frac{3}{4}+\frac{3}{4}x29+\frac{75}{100}x30+\frac{75}{100}x40\)

\(=\frac{3}{4}+\frac{3}{4}x29+\frac{3}{4}x30+\frac{3}{4}x40\)

\(=\frac{3}{4}x\left(1+29+30+40\right)\)

\(=\frac{3}{4}x100\)

  = 75

75/100 + 3/4 x 29 + 75% x 30 + 0,75 x 40

=75/100 + 3/4 x 29 + 75/100 x 30 + 75/100 x 40

=3/4 x 1 +3/4 x 29 + 3/4 x 30 + 3/4 x 40

=3/4 x ( 1+29+30+40 )

=3/4 x         100

=75/100 x 100

=0,75 x 100

=75

_K nha~

#bối#

-----------Hok Tốt-----------

\(\frac{75}{100}\)+ \(\frac{3}{4}\)x 29 + 75% x 30 + 0,75 x 40

\(\Leftrightarrow\)\(\frac{3}{4}+\frac{3}{4}\)x 29 + \(\frac{3}{4}\)x 30 + \(\frac{3}{4}\)x 40

\(=\frac{3}{4}.\left(29+1+30+40\right)\)

\(=\frac{3}{4}.100\)

\(=0,75.100=75\)

`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x={-3/2; 1/2}.`