c: \(\dfrac{x+1}{-5}=\dfrac{-20}{x+1}\)(Điều kiện: \(x\ne-1\))

=>\(\left(x+1\right)^2=\left(-20\right)\cdot\left(-5\right)=100\)

=>\(\left[{}\begin{matrix}x+1=10\\x+1=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=9\left(nhận\right)\\x=-11\left(nhận\right)\end{matrix}\right.\)

d: \(\dfrac{-4}{5}-\left(x+\dfrac{1}{2}\right)=\dfrac{1}{-5}-\dfrac{15}{10}\)

=>\(\dfrac{-4}{5}-x-\dfrac{1}{2}=\dfrac{-1}{5}-\dfrac{15}{10}\)

=>\(\dfrac{-13}{10}-x=\dfrac{-17}{10}\)

=>\(x=\dfrac{-13}{10}+\dfrac{17}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

d:

ĐKXĐ: x<>-1

 \(-\dfrac{195}{13}=\dfrac{30}{x+1}=\dfrac{y^2}{-15}\)

=>\(\dfrac{30}{x+1}=\dfrac{y^2}{-15}=-15\)

=>\(\left\{{}\begin{matrix}x+1=\dfrac{30}{-15}=-2\\y^2=\left(-15\right)\cdot\left(-15\right)=225\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\left(nhận\right)\\y\in\left\{15;-15\right\}\end{matrix}\right.\)

e: \(x+\dfrac{1}{5}=\dfrac{4}{-10}-\dfrac{3}{2}\)

=>\(x+\dfrac{1}{5}=\dfrac{-2}{5}-\dfrac{3}{2}\)

=>\(x=\dfrac{-2}{5}-\dfrac{3}{2}-\dfrac{1}{5}=\dfrac{-3}{2}-\dfrac{3}{5}=\dfrac{-21}{10}\)

c.

\(\dfrac{x+1}{-5}=\dfrac{-20}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=-20.\left(-5\right)\)

\(\Leftrightarrow\left(x+1\right)^2=100\)

\(\Rightarrow x+1=10\) hoặc \(x+1=-10\)

\(\Rightarrow x=9\) hoặc \(x=-11\)

d.

\(-\dfrac{4}{5}-\left(x+\dfrac{1}{2}\right)=\dfrac{1}{-5}-\dfrac{15}{10}\)

\(\Rightarrow-\dfrac{4}{5}-\left(x+\dfrac{1}{2}\right)=-\dfrac{17}{10}\)

\(\Rightarrow x+\dfrac{1}{2}=-\dfrac{4}{5}+\dfrac{17}{10}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{9}{10}\)

\(\Rightarrow x=\dfrac{9}{10}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{2}{5}\)

d.

\(\dfrac{-195}{13}=\dfrac{30}{x+1}=\dfrac{y^2}{-15}\)

\(\Rightarrow-15=\dfrac{30}{x+1}=\dfrac{y^2}{-15}\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=30:\left(-15\right)\\y^2=-15.\left(-15\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=-2\\y^2=15^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=\pm15\end{matrix}\right.\)

e.

\(x+\dfrac{1}{5}=\dfrac{4}{-10}-\dfrac{3}{2}\)

\(x+\dfrac{1}{5}=-\dfrac{2}{5}-\dfrac{3}{2}\)

\(x+\dfrac{1}{5}=-\dfrac{19}{10}\)

\(x=-\dfrac{19}{10}-\dfrac{1}{5}\)

\(x=-\dfrac{21}{10}\)

P= {a;b;c;d}

Mình nghĩ thế

bạn để lại màn hình là được

chữ xấu vãiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

a) Để C cõ nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0< =>x\ne1\\2-2x^2\ne0< =>x\ne\pm1\end{matrix}\right.\)

b) C = \(\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2x^2-2}=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

= \(\dfrac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x-x^2-1}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{2\left(x+1\right)}\)

c) Để C = \(\dfrac{-1}{2}\)

<=> \(\dfrac{1}{2\left(x+1\right)}=\dfrac{-1}{2}\)

<=> x+ 1 = -1

<=> x = -2 (tm)

d) Để C nguyên

<=> \(\dfrac{1}{2\left(x+1\right)}\) nguyên

<=> 2(x+1) có dạng \(\dfrac{1}{k}\left(k\in Z,k\ne0\right)\)

<=> \(x+1=\dfrac{1}{2k}< =>x=\dfrac{1}{2k}-1\)

Bài 39:

a: Đặt \(\dfrac{5}{x-2}-\dfrac{1}{x+2}+\dfrac{4}{\left(x-2\right)\left(x+2\right)}=0\)

=>5x+10-x+2+4=0

=>4x+16=0

hay x=-4(nhận)

b: Đặt \(\dfrac{-2}{x^2-x+1}+x+1=0\)

\(\Leftrightarrow-2+x^3+1=0\)

\(\Leftrightarrow x^3=1\)

hay x=1(nhận)

c, Thay x = -1/2 ; y = 2/3 vào ta được 

\(P=\dfrac{2.1}{4}+3\left(-\dfrac{1}{2}\right).\dfrac{2}{3}+\dfrac{4}{9}=\dfrac{1}{2}-1+\dfrac{4}{9}=-\dfrac{1}{18}\)

d, Thay a = -1/3 ; b = -1/6 ta được 

\(D=12\left(-\dfrac{1}{3}\right).\dfrac{1}{36}=\dfrac{12}{-3.12.3}=-\dfrac{1}{9}\)

 ĐỂ THOÁT KHỎI PHẦN MỀM THE MONKEY EYES  , EM NHẤN PHÍM 

A.SHIFT        B.TAD          C.ESC          D. ENTER

Đáp án là C

Chúc bạn học tốt!! ^^ 

Chọn C

Xét ΔKIH có MD//IH

nên IM/IK=HD/HK

Xét ΔHKI có DN//KI

nên IN/IH=KD/KH

IM/IK+IN/ID

=HD/HK+DK/KH

=HK/HK=1