Question

[Ngan Phan Thi]

Câu 1: Tìm giá trị nhỏ nhất của biểu thức sau:

H=↑x-3↑+2020; B=(x-1)²+2021

Câu 2: Cho 3a²-b²/a²+b²=3/4. Tính a²/b²

Câu 3: Cho ab=c². CMR a²+c²/b²+c²=a/b

Câu 4: Cho a/2=b/5=c/7. Tìm giá trị của biểu thức A=a-b+c/a+2b-c

Answer 1

[Nguyễn Giang]

\(\text{Câu 1: }\)

\(a,\left|x-3\right|\ge0\left(\forall x\in N\right)\)

\(\Rightarrow\left|x-3\right|+2020\ge2020\left(\forall x\in N\right)\)

\(\text{Dấu}"="\text{xảy ra}\Leftrightarrow\left|x-3\right|+2020=2020\)

\(\Leftrightarrow\left|x-3\right|=2020-2020\)

\(\Leftrightarrow\left|x-3\right|=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=0+3\)

\(\Leftrightarrow x=3\) \(\text{Vậy }x=3\text{ để H có GTNN}\)

\(b,\left(x-1\right)^2\ge0\left(\forall x\in N\right)\)

\(\Rightarrow\left(x-1\right)^2+2021\ge2021\left(\forall x\in N\right)\)

\(\text{Dấu}"="\text{xảy ra}\Leftrightarrow\left(x-1\right)^2+2021=2021\)

\(\Leftrightarrow\left(x-1\right)^2=2021-2021\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=0+1\)

\(\Leftrightarrow x=1\) \(\text{Vậy }x=1\text{ để B có GTNN}\)

\(\text{Câu 2:}\)

\(\frac{3a^2-b^2}{a^2+b^2}=\frac{3}{4}\)

\(\Rightarrow\left(3a^2-b^2\right).4=\left(a^2+b^2\right).3\)

\(\Rightarrow12a^2-4b^2=3a^2+3b^2\)

\(\Rightarrow12a^2-3a^2=3b^2+4b^2\left(\text{quy tắc chuyển vế}\right)\)

\(\Rightarrow a^2.\left(12-3\right)=b^2.\left(3+4\right)\)

\(\Rightarrow a^2.9=b^2.7\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{7}{9}\left(\text{tính chất của tỉ lệ thức}\right)\)

\(\text{Câu 3:}\)

\(ab=c^2;\frac{a^2+c^2}{b^2+c^2}\left(1\right)\)

\(\text{Thay }c^2=ab\text{ vào }\left(1\right)\)

\(\Rightarrow\frac{a^2+ab}{b^2+ab}=\frac{a.\left(a+b\right)}{b.\left(a+b\right)}=\frac{a}{b}\left(2\right)\)

\(\text{Từ (1) và (2)}\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\left(đpcm\right)\)

\(\text{Câu 4: }\)

\(A=\frac{a-b+c}{a+2b-c}\)

\(\frac{a}{2}=\frac{b}{5}\Rightarrow a=\frac{2}{5}.b;\frac{c}{7}=\frac{b}{5}\Rightarrow c=\frac{7}{5}.b\)

\(\text{Thay }a=\frac{2}{5}.b;c=\frac{7}{5}.b\text{ vào }A\)

\(\Rightarrow A=\frac{\frac{2}{5}.b-b+\frac{7}{5}.b}{\frac{2}{5}.b+2b-\frac{7}{5}.b}=\frac{b.\left(\frac{2}{5}-1+\frac{7}{.5}\right)}{b.\left(\frac{2}{5}+2-\frac{7}{5}\right)}=\frac{\frac{2}{5}-\frac{5}{5}+\frac{7}{5}}{\frac{2}{5}+\frac{10}{5}-\frac{7}{5}}=\frac{\frac{2-5+7}{5}}{\frac{2+10-7}{5}}=\frac{4}{5}:1=\frac{4}{5}\)

\(\text{Vậy }A=\frac{4}{5}\)