Đăng tách ra mỗi câu 1 lần.

Bài 7:

Ta có: f(x)=4

\(\Leftrightarrow x-4-3\left(x+1\right)=4\)

\(\Leftrightarrow x-4-3x-3-4=0\)

\(\Leftrightarrow-2x=11\)

hay \(x=-\dfrac{11}{2}\)

mn xa lánh em quá hic:((((((

giúp em với mai em nộp bài đó ạ;-;

\(1,=\left(x-3\right)^2\\ 2,=\left(5+x\right)^2\\ 3,=\left(\dfrac{1}{2}x+2b\right)^2\\ 4,=\left(\dfrac{1}{3}-y^4\right)^2\\ 5,=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ 6,=\left(2y-5\right)\left(4y^2+10y+25\right)\\ 7,=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\\ 8,=\left(x-5\right)^2\\ 9,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)

Bn dùng hằng đẳng thức đáng nhớ nhé.

Thêm một chút kiên thức về bài:

\(\left(\dfrac{a}{b}\right)^2=\dfrac{a^2}{b^2}\)

\(a^{xy}=\left(a^x\right)^y\)

\(8,=\left(3x+2-2\right)\left(3x+2+2\right)=3x\left(3x+4\right)\\ 9,=\left(2x-5y\right)\left(2x+5y\right)\\ 10,=\left(2x-7\right)\left(2x+7\right)\\ 11,=\left(2z+3\right)\left(4z^2-6z+9\right)\\ 12,=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\\ 13,=\left(x^{16}-1\right)\left(x^{16}+1\right)=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ 14,=\left(2x+1\right)^2\\ 15,=\left(x-10\right)^2\\ 16,=\left(y^2-7\right)^2\)

\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)

\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)

\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Đánh giá giúp nhá cảm ơn

Bài 7:

a: \(A=x+\sqrt{x}\ge0\forall x\)

Dấu '=' xảy ra khi x=0

\(a=\dfrac{1}{9}.\left(999...9\right)=\dfrac{1}{9}.\left(100...0-1\right)=\dfrac{1}{9}\left(10^n-1\right)\)

\(b=100...0+5=10^n+5\)

\(\Rightarrow ab+1=\dfrac{1}{9}\left(10^n-1\right)\left(10^n+5\right)+1=\dfrac{1}{9}\left(10^{2n}+4.10^n+4\right)=\dfrac{1}{9}\left(10^n+2\right)^2\)

\(=\left(\dfrac{10^n+2}{3}\right)^2\)

Ta có: \(10\equiv1\left(mod3\right)\Rightarrow10^n\equiv1\left(mod3\right)\)

\(\Rightarrow10^n+2⋮3\)

\(\Rightarrow\dfrac{10^n+2}{3}\in Z\)

\(\Rightarrow\left(\dfrac{10^n+2}{3}\right)^2\) là SCP hay \(ab+1\) là SCP

bài toán là gì đọc đề bài tui giải cho tui học lớp 12

nhìn chóng hết cả mặt chắc mình trẻ quá nên mắt kém

bài nào 

tui đội tuyển Toán 8 nè