a. ta có :

\(\hept{\begin{cases}\left|x-1\right|+\left|x-4\right|\ge\left|x-1-x+4\right|=3\\\left|x-2\right|+\left|x-3\right|\ge\left|x-2-x+3\right|=1\\\left|2x-5\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\)thay lại thấy thỏa mãn . Vậy x=5/2 là nghiệm

b.ta có 

\(\hept{\begin{cases}\left|x+1\right|+\left|x-1\right|\ge\left|x+1-x+1\right|=2\\\left|x+2\right|+\left|x-5\right|\ge\left|x+2-x+5\right|=7\\\left|3x+2\right|\ge0\end{cases}}\)

Vậy phương trình ban đầu có nghiệm \(\Rightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\)thay lại thấy thỏa mãn . Vậy x=-2/3 là nghiệm

Đáp án :

\(x\in\varnothing\)

# Hok tốt !

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Gợi ý thôi còn bn tự làm

Xét 4 TH : \(x\ge\frac{7}{2};1\le x< \frac{7}{2};-5\le x< 1;x< -5\)

\(13^{\left(x-2\right)\left(2x-5\right)}=1=13^0\)

\(\Rightarrow\left(x-2\right)\left(2x-5\right)=0\Leftrightarrow x=2;x=\frac{5}{2}\)

Ta có với mọi \(a\in Z\)thì \(a^0=1\)

\(\Rightarrow13^{\left(x-2\right)\left(2x-5\right)}=13^0=1\)

\(\Rightarrow\left(x-2\right)\left(2x-5\right)=0\)

\(\Rightarrow x-2=0\)hoặc \(2x-5=0\)

\(TH1:\)\(x-2=0\)

\(\Rightarrow x=2\)

\(TH2:\)\(2x-5=0\)

\(\Rightarrow2x=5\)

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

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

Trả lời:

\(13^{\left(x-2\right)\left(2x-5\right)}=1\)

\(\Leftrightarrow13^{\left(x-2\right)\left(2x-5\right)}=13^0\)

\(\Rightarrow\left(x-2\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{2}\end{cases}}}\)

Vậy x = 2; x = 5/2

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).