Bài 4:

a: Bảng giá trị:

Vẽ đồ thị:

b: Để đồ thị hàm số y=2(m-1)x-2 cắt đồ thị hàm số y=2x+3 thì 2(m-1)<>2

=>m-1<>1

=>m<>2

Bài 3:

\(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}+\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

\(=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}+1+x-\sqrt{x}+1}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\frac{2x+2}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(x+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right)\cdot\frac{a-2\sqrt{a}+1}{\sqrt{a}+1}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

Mk lm đc ý a,B thôi

5 đc ko?

là sao ko hiểu lắm

Theo hình, ta có: \(DG=\frac25\times DC\)

=>\(DG=\frac25\times AB\)

Vì AB//DG

nên \(\frac{EA}{EG}=\frac{EB}{ED}=\frac{AB}{DG}=\frac52\)

\(\frac{ED}{EB}=\frac25\)

=>\(\frac{DE}{DB}=\frac27\)

=>\(S_{ADE}=\frac25\times S_{ABD}\)

=>\(S_{ABD}=96:\frac25=96\times\frac52=48\times5=240\left(\operatorname{cm}^2\right)\)

ABCD là hình chữ nhật

=>\(S_{ABCD}=2\times S_{BAD}=2\times336=672\left(\operatorname{cm}^2\right)\)

Ta có: \(\frac{ED}{EB}=\frac25\)

=>\(\frac{S_{AED}}{S_{AEB}}=\frac25\)

=>\(\frac{96}{S_{AEB}}=\frac27\)

=>\(S_{AEB}=96\times\frac72=96\times3,5=336\left(\operatorname{cm}^2\right)\)

Vì \(\frac{EA}{EG}=\frac52\)

nên \(\frac{S_{DEA}}{S_{DEG}}=\frac52\)

=>\(\frac{96}{S_{DEG}}=\frac52\)

=>\(S_{DEG}=96\times\frac25=38,4\operatorname{\left(cm\right)}^2\)

Ta có: \(S_{DEG}+S_{BEGC}=S_{BCD}\)

=>\(S_{BEGC}=240-38,4=201,6\left(cm^2\right)\)