a)

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\(\Leftrightarrow\widehat{A}+\widehat{D}=180^o\)

\(\Leftrightarrow\widehat{D}=180^o-120^o\)

\(\Leftrightarrow\widehat{D}=60^o\)

b)

Góc D/B = \(\frac{1}{2}\)

\(\Leftrightarrow\widehat{B}=120^o\)

\(\widehat{C}=180^o-\widehat{D}=60^o\)

giúp mik please :3

( 2x - 1 )³ - 2x ( 4x² - 3x ) = 5

<=> 8x³ - 12x² + 6x - 1 - 8x³ + 6x² - 5 = 0

<=> - 6x² + 6x - 6 = 0

<=> x² - x + 1 = 0

<=> ( x - 1/2 )² = - 3/4 ( kh đúng )

=> pt vô nghiệm

\(\left(2x-1\right)^3-2x\left(4x^2-3x\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+6x^2=5\)

\(\Leftrightarrow-6x^2+6x-6=0\Leftrightarrow x^2-x+1=0\)

Ta có : \(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

Vậy phương trình vô nghiệm 

\(4x^2-4x\left(x-3\right)+\left(3-x\right)^2=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x\left(x-3\right)+\left(x-3\right)^2=0\)

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

Từ:

 \(\hept{\begin{cases}a^2\left(b+c\right)=2012\\b^2\left(a+c\right)=2012\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a^2\left(b+c\right)-b^2\left(a+c\right)=0\\b^2\left(a+c\right)+a^2\left(b+c\right)=4024\end{cases}}\)

\(a^2\left(b+c\right)-b^2\left(a+c\right)=0\)

\(\Leftrightarrow ab\left(a-b\right)+c\left(a+b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow ab+bc+ca=0\left(a\ne b\right)\)

\(b^2\left(a+c\right)+a^2\left(b+c\right)=4024\)

\(\Leftrightarrow ab\left(a+b\right)+c\left(a+b\right)^2-2abc=4024\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)-2abc=4024\)

\(\Leftrightarrow2abc=-4024\)

\(\Leftrightarrow abc=-2012\)

\(ab\left(a-b\right)+c\left(a+b\right)\left(a-b\right)=0\)

\(\Leftrightarrow c\left(a+b\right)\left(a-b\right)=-ab\left(a-b\right)\)

\(\Leftrightarrow c\left(a+b\right)=-ab\)

\(\Leftrightarrow c^2\left(a+b\right)=-abc=2012\)

(x - 2)³ - x³ + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x³ + 6x² = 5

<=> 12x - 8 = 5

<=> 12x = 13

<=> \(x=\frac{13}{12}\)

NHA

MÌNH MÁY TÍNH  KO RÕ NÊN KO THẤY CÂU HỎI CỦA YOU

| x + 3 | = 2x - 1

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

\(\Rightarrow\orbr{\begin{cases}2x-x=3+1\\x+2x=1-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\3x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=\frac{-2}{3}\end{cases}}\)

Cách đó chỉ giải với 2 vế là GTTĐ thôi nhá @Sơn 

CTTQ : \(\left|f\left(x\right)\right|=g\left(x\right)\)ĐK : \(g\left(x\right)\ge0\)

TH1 : \(f\left(x\right)=g\left(x\right)\)

TH2 : \(f\left(x\right)=-g\left(x\right)\)

-> rồi xét x có thỏa mãn với đk ko rồi kết luận 

\(\left|x+3\right|=2x-1\)ĐK : \(x\ge\frac{1}{2}\)

TH1 : \(x+3=2x-1\Leftrightarrow x=4\)(tmđk )

TH2 : \(x+3=1-2x\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)(ktmđk) 

Vậy x = 4