\(\left|4-3x\right|-x=1\)

Trường hợp 1: \(4-3x\ge0\Rightarrow x\le\frac{4}{3}\)

\(4-3x-x=1\Rightarrow4-4x=1\Rightarrow4\left(1-x\right)=1\Rightarrow1-x=\frac{1}{4}\Rightarrow x=\frac{3}{4}\) (Thoả mãn)

Trường hợp 2: \(4-3x\le0\Rightarrow x\ge\frac{4}{3}\)

\(-4+3x-x=1\Rightarrow-4+2x=1\Rightarrow2\left(x-2\right)=1\Rightarrow x-2=\frac{1}{2}\Rightarrow x=\frac{5}{2}\) (Thoả mãn)

Vậy \(x\in\left\{\frac{3}{4};\frac{5}{2}\right\}\)

b4: a đúng ; b sai ; c đúng

b5:

(viết B₂ ở dưới B₁ cho mình nha)

có a⊥b;b⊥c => a // b

Có ∠A₁ = ∠B₂ (t/c hai đường thẳng song song)

  => ∠B₂ = 115^o 

Mà ∠B₁ + ∠B₂ = 180^o 

  => ∠B₁ = 180^o - ∠B₂ 

               = 180^o - 115^o 

       => ∠B₁ = 65^o 

\(25^{3x}.125^4=\frac{1}{25}\)

\(\Rightarrow\left(5^2\right)^{3x}.\left(5^3\right)^4=\frac{1}{5^2}\)

\(\Rightarrow5^{6x}.5^{12}=\frac{1}{5^2}\)

\(\Rightarrow5^{6x+12}=5^{-2}\)

=> 6x + 12 = -2

=> 6x = -14

\(\Rightarrow x=\frac{-7}{3}\)

x = \(\left(-\frac{4}{6}\right)+\frac{2}{3}\)

x = 0

# hok tốt !

TL:

\(X\)\(=\)\(\left(-\frac{4}{6}\right)\)\(+\frac{2}{3}\)

\(X=0.\)

\(~HT~\)

Bài 72 :

a) Vì \(\left|2x-\frac{1}{3}\right|\ge0\forall x\)

\(\Rightarrow\left|2x-\frac{1}{3}\right|-\frac{7}{4}\ge-\frac{7}{4}\forall x\)

\(\Rightarrow A\ge-\frac{7}{4}\)

Dấu "=" xảy ra <=> \(\left|2x-\frac{1}{3}\right|=0\)

\(\Leftrightarrow2x-\frac{1}{3}=0\Leftrightarrow2x=\frac{1}{3}\Leftrightarrow x=\frac{1}{6}\)

b) Vì \(\hept{\begin{cases}\frac{1}{3}\left|x-2\right|\ge0∀x\\2\left|3-\frac{1}{2}y\right|\ge0∀y\end{cases}}\)\(\Rightarrow\frac{1}{3}\left|x-2\right|+2\left|3-\frac{1}{2}y\right|+4\ge4∀y\)

=> B ≥ 4

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x-2\right|=0\\\left|3-\frac{1}{2}y\right|=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-2=0\\3-\frac{1}{2}y=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\\frac{1}{2}y=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=6\end{cases}}\)

Bài 73 :

a) Vì | 1 + 2x | ≥ 0 ∀ x

\(\Rightarrow\frac{1}{4}\left|1+2x\right|\ge0\forall x\)

\(\Rightarrow2,25-\frac{1}{4}\left|1+2x\right|\le2,25\forall x\)

=> A ≤ 2,25

Dấu "=" xảy ra <=> | 1 + 2x | = 0 <=> 1 + 2x = 0 <=> 2x = -1 <=> \(x=\frac{-1}{2}\)

b) Vì | 2x - 3 | ≥ 0 ∀ x

\(\Rightarrow\frac{1}{2}\left|2x-3\right|\ge0\forall x\)

\(\Rightarrow3+\frac{1}{2}\left|2x-3\right|\ge3\forall x\)

\(\Rightarrow\frac{1}{3+\frac{1}{2}\left|2x-3\right|}\le\frac{1}{3}\forall x\)

\(\Rightarrow B\le\frac{1}{3}\)

Dấu "=" xảy ra <=> | 2x - 3 | = 0 <=> 2x - 3 = 0 <=> 2x = 3 \(\Leftrightarrow x=\frac{3}{2}\)