a) \(x+\left|x-2\right|=7\)

\(\Leftrightarrow\left|x-2\right|=7-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=7-x\\x-2=-7+x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=9\\0x=-5\left(loại\right)\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{9}{2}\)

b) \(\left|x-3\right|+\left|x-5\right|=9\left(1\right)\)

Ta thấy :

\(\left|x-3\right|+\left|x-5\right|\ge\left|x-3+x-5\right|=\left|2x-8\right|\)

\(pt\left(1\right)\Leftrightarrow\left|2x-8\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=9\\2x-8=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c) \(\left|x-1\right|+\left|x+1\right|=10\left(1\right)\)

Ta thấy :

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

\(pt\left(1\right)\Leftrightarrow\left|2x\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

a) \(x+\left|x-2\right|=7\)

\(\Rightarrow\left\{{}\begin{matrix}x+\left(x-2\right)=7\left(x\ge2\right)\\x-\left(x-2\right)=7\left(x< 2\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+x-2=7\\x-x+2=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x-2=7\\2=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=9\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow2x=-9\)

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

\(5^{x-3}=25^9\)

\(\Rightarrow5^{x-3}=\left(5^2\right)^9\)

\(\Rightarrow5^{x-3}=5^{18}\)

\(\Rightarrow x-3=18\)

\(\Rightarrow x=18+3\)

\(\Rightarrow x=21\)

Vậy: x=21

\(5x^2=1620\)

\(\Rightarrow x^2=1620:5\)

\(\Rightarrow x^2=324\)

\(\Rightarrow x^2=18^2\)

\(\Rightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)

Vậy: x=18 hoặc x=-18

\(5^x+10x=625+10x\)

\(\Rightarrow5^x=625+10x-10x\)

\(\Rightarrow5^x=625\)

\(\Rightarrow5^x=5^4\)

\(\Rightarrow x=4\)

Vậy x=4

\(2^x:8=4\)

\(\Rightarrow2^x=4\cdot8\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(x=5\)

Vậy: x=5

\(2^x:8=4\)

\(\Rightarrow2^x=4.8\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^5=32\)

\(\Rightarrow x=5\)

A = 1 + 3 + 5 + ... + 299

a) Số số hạng của A:

(299 - 1) : 2 + 1 = 150 (số)

A = (299 + 1) . 150 : 2 = 22500

b) Số hạng thứ 74 của A:

1 + 73 × 2 = 147

= 7 - 3 căn 5 + 1

= 

Nó ra xấp xỉ mà nhỉ đề vô lí vậy ta

Giải thích seo cho hs lớp 7 hiểu cùng ạ.

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

\(x=\dfrac{-1}{3}\)

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3

\(\sqrt[]{x+2}=-100\)

vì \(\sqrt[]{x+2}\ge0\)

Nên phương trình trên vô nghiệm

$\sqrt{x+2}=-100$

vì $\sqrt{x+2}\ge 0$

Nên phương trình trên vô nghiệm

Chúc bạn nha

\(A=\left(x-\dfrac{1}{4}\right)^4+\left|x-2y\right|+1\)

vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^4\ge0,\forall x\\\left|x-2y\right|\ge0,\forall x;y\end{matrix}\right.\)

\(\Rightarrow A=\left(x-\dfrac{1}{4}\right)^4+\left|x-2y\right|+1\ge1\)

Dấu "=" xảy ra khi và chỉ khi

\(\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\x-2y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{x}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{8}\end{matrix}\right.\)

Vậy \(GTNN\left(A\right)=1\left(tạix=\dfrac{1}{4};y=\dfrac{1}{8}\right)\)

Ta có:

(x - 1/4)⁴ ≥ 0 với mọi x ∈ R

(x - 2y)² ≥ 0 với mọi x, y ∈ R

(x - 1/4)⁴ + (x - 2y)² ≥ 0 với mọi x, y ∈ R

(x - 1/4)⁴ + (x - 2y)² + 1 ≥ 1 với mọi x, y ∈ R

Vậy GTNN của A là 1 khi x = 1/4 và y = 1/8