a, |x-2|<7
b,|5+x|>4
giúp mk vs mk tick cho :)))))
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\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-2-5\left(x+2\right)-2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x-2-5x-10-2x-3=0\)
\(\Leftrightarrow-6x-15=0\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)
Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đkxđ : x khác 2 , x khác -2.
<=> \(\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{x^2-4}=0\)
<=> \(\dfrac{1.\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=>\(x-2-5x-10-2x+3=0\)
<=> \(-6x-9=0\)
<=> \(x=-\dfrac{9}{6}=-\dfrac{3}{2}\left(nhận\right)\)
Vậy pt có nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
⇔ \(\dfrac{x-7}{7}=\dfrac{y-6}{6}\)
⇔ \(\dfrac{x}{7}=\dfrac{y}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{x-y}{7-6}=-4\)
⇒ \(\left\{{}\begin{matrix}x=-28\\y=-24\end{matrix}\right.\)
Vậy ...
7,14 + 7,14 + 7,14 x 9 - 7,14
= 7,14 x 1 + 7,14 x 1 + 7,14 x 9 - 7,14 x 1 )
= 7,14 x ( 1 + 1 + 9 - 1 )
= 7,14 x 10
= 71,4
0,5 x 9,6 x 2
= ( 0,5 x 2 ) x 9,6
= 1 x 9,6
= 9,6
A)7,14×7,14+7,14×9-7,14=7,14×(1+1+9-1)
=7,14×10
=71,4
B)0,5×9,6×2=(0,5×2)×9,6=1×9,6
a, x^2 - 2x + 7
= x( x-2) + 7
ta có x(x-2) chia hết cho x- 2
nên để x^2 - 2x + 7 chia hết cho 2
thì 7 chia hết cho x- 2
=> x-2 thuộc ước của 7
đến đây tự làm tiếp
Ta có 2A=\(2^2+2^3+...+2^{101}\)
=>2A-A=A=\(\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
=> A= \(2^{101}-2\)
Mà \(A+1=2^x\)
=> \(2^x=2^{101}-2^0\)
Bạn xem lại đề nhé mk cx ko rõ nữa
2A=\(2\left(2+2^2+2^3+....+2^{100}\right)\)
2A=\(2^2+2^3+2^4+.....+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+...2^{101}\right)-\left(2+2^2+2^3+....+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
Vậy A= \(2^{101}-2\)
\(a,\frac{7}{x}=\frac{x}{28}=>x\cdot x=28\cdot7=>x^2=196=>x^2=14^2\)\(=>x=14\)
\(b,\frac{10+x}{x+17}=\frac{3}{4}=>\left(10+x\right)\cdot4=\left(x+17\right)\cdot3=>40+x4=x3+51\)\(=>x4-x3=51-40=>x=11\)
A)\(\left|x-2\right|< 7\Rightarrow\left|x-2\right|\in\left\{0;1;2;3;4;5;6\right\}\)
\(\Rightarrow\left(x-2\right)\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6\right\}\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8\right\}\)
B) \(\left|5+x\right|>4\Rightarrow\left|5+x\right|\in\left\{5;6;7;8;9;...\right\}\)
\(\Rightarrow\left(5+x\right)\in\left\{...;-7;-6;-5;5;6;7;8;...\right\}\)
\(\Rightarrow x\in\left\{...;-5;-4;-3;-2;-1;0;1;2;3;4;5;...\right\}\)