-(-30)-(-x)=-(-13)

30+x=13

=>x=13-30

=>x=-17

Ta có : \(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)642-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)

=> \(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)642-0,04:\left(x+1,05\right):0,12=19\)

=> 

(nãy bấm nhầm) tiếp nà :

=> \(\left(\frac{1}{11}-\frac{1}{21}\right)462-0,04:\left(x+1,05\right):0,12=19\)

=> \(\frac{10}{231}.462-0,04:\left(x+1,05\right):0,12=19\)

=> \(20-0,04:\left(x+1,05\right):0,12=19\)

=> 0,04 : (x + 1,05) : 0,12 = 1

=> 0,04 : (x + 1,05) = 0,12

=> \(x+1,05=\frac{1}{3}\) 

=> \(x=\frac{1}{3}-1,05=...\)

\(\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)562-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)

\(\Rightarrow2.\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)

\(\Rightarrow\left(\frac{1}{11}-\frac{1}{21}\right)462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)

\(\Rightarrow\frac{10}{231}.462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)

\(\Rightarrow20-\left[0,04:\left(x+1,05\right)\right]0,12=19\)

\(\Rightarrow\left[0,04:\left(x+1,05\right)\right]:0,12=20-19=1\)

\(\Rightarrow0,04:\left(x+1,05\right)=1.0,12=0,12\)

\(\Rightarrow x+1,05=0,04:0,12=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-1,05=\frac{1}{3}-\frac{105}{100}=-\frac{43}{60}\)

(Làm vậy đúng không?)

c,11-(-53+x)=97

-53+x=11-97

-53+x= -86

x = -86-(-53)

x= -33

Câu hỏi của Mai Chi - Toán lớp 7 - Học toán với OnlineMath

a)\(\left(2x-3\right)\left(x+1\right)< 0\)

\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\)  hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại)  hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)

\(\Leftrightarrow-1< x< \frac{3}{2}\)

b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)

c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)

Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow x=4\)

đề câu c sai rầu kìa má

a) |-x + 2| = -|y + 9|

=> |-x + 2| + |y + 9| = 0

Ta có: |-x + 2| \(\ge\)0 \(\forall\)x

|y + 9| \(\ge\)0 \(\forall\)y

=> |-x + 2| + |y + 9| \(\ge\)0 \(\forall\)x; y

Dấu "=" xảy ra khi : \(\hept{\begin{cases}-x+2=0\\y+9=0\end{cases}}\) => \(\hept{\begin{cases}x=2\\y=-9\end{cases}}\)

Vậy ...

b) |3x + 4| + |2y - 10| \(\le\)0

Ta có: |3x +  4| \(\ge\)0 \(\forall\)x

        |2y - 10| \(\ge\)0 \(\forall\)y

=> |3x + 4| + |2y - 10| \(\ge\) 0 \(\forall\)x;y

Dấu "=" xảy ra khi : \(\hept{\begin{cases}3x+4=0\\2y-10=0\end{cases}}\) <=> \(\hept{\begin{cases}3x=-4\\2y=10\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{4}{3}\\y=5\end{cases}}\)

vậy ...

c) |-x - 3| + |y + 7| < 0

Ta có: |-x - 3| \(\ge\)0 \(\forall\)x

      |y + 7| \(\ge\)0 \(\forall\)y

=> |-x - 3| + |y + 7| \(\ge\)0 \(\forall\)x; y

=> ko có giá trị x, y thõa mãn đb

Lời giải của mình ở đây nha bạn!

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\(S=\left\{\frac{4023}{2};\frac{4015}{2}\right\}\)