Lời giải:

Ta có:\(F(x)=\int (2x-3)\ln xdx\)

Đặt \(\left\{\begin{matrix} u=\ln x\\ dv=(2x-3)dx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\int (2x-3)dx=x^2-3x\end{matrix}\right.\)

Do đó:

\(F(x)=\int (2x-3)\ln xdx=(x^2-3x)\ln x-\int (x^2-3x).\frac{dx}{x}\)

\(=(x^2-3x)\ln x-\int (x-3)dx=(x^2-3x)\ln x-(\frac{x^2}{2}-3x)+c\)

Với \(x=1\)

\(F(1)=\frac{5}{2}+c=0\Rightarrow c=\frac{-5}{2}\)

Vậy \(F(x)=(x^2-3x)\ln x-\frac{x^2}{2}+3x-\frac{5}{2}\)

\(\Rightarrow 2F(x)+x^2-6x+5=2(x^2-3x)\ln x-x^2+6x-5+x^2-6x+5\)

\(=2(x^2-3x)\ln x=0\)

\(\Leftrightarrow \left[\begin{matrix} x=0\\ x=3\\ x=1\end{matrix}\right.\)

Tức là pt có 3 nghiệm.

a: Thay x=1 và y=-3 vào y=(m-1)x, ta được:

m-1=-3

hay m=-2

b: f(x)=-3x

f(2/3)=-2

f(-4)=12

c:f(-1)=3 nên M thuộc đồ thị

f(6)=-18<>-9 nên N không thuộc đồ thị

xét từng th

A(0;-2)=>0^2-2=0-2=-2=>A THUỘC HÀM SỐ Y=F(X)

B(-1;-3)=>-1^2-2=-1-2=-3=>B THUỘC HÀM SỐ Y=F(X)

C(1;-3)=>1^2-2=2-2=0=>C KO THUỘC HÀM SO Y=F(X)

=>NHỮNG ĐIỂM THUỘC H/S Y=F(X)LÀ ĐIỂM A(0;-2);B(-1;-3)

\(f\left(x\right)=2x.\)

a) Thay \(x=-2\) vào \(f\left(x\right)\) ta được:

\(f\left(-2\right)=2.\left(-2\right)\)

\(f\left(-2\right)=-4.\)

+ Thay \(x=2\) vào \(f\left(x\right)\) ta được:

\(f\left(2\right)=2.2\)

\(f\left(2\right)=4.\)

Chúc bạn học tốt!

đặt y=f(x)=2x

ta có:

f(-2)=2.(-2)=-4

f(2)=2.2=4

mình ko vẽ hình nhé

Câu C

Đúng nhất !!!

f(x)=-2x

a: f(-2)=4

f(4)=-8

c: f(2)=-4 nên A không thuộc đồ thị

f(-3)=6 nên B thuộc đồ thị

f(-1/2)=1 nên C thuộc đồ thị

Ta có: f(-1)=3*(-1)-7=-3-7=-10

          f(0)=3*0-7=0-7=-7

Vậy f(-1)=-10  ;f(0)=-7

Xét điểm M(2;-1) ta có

Với x=2 => y=3*2-7=-1

=> Điểm M(2;-1) thuộc đồ thị hàm số y=3x-7

Vậy điểm M(2;-1) thuộc đồ thị hàm số y=3x-7

Tích mk nha pn....

Ta có :y=f(x)=f(-1)=3*-1-7=-10

y=f(x)=f(0)=3*0-7=-7

M(2;-1)=>x=2;y=-1

=>y=3*2-7=-1

=>M thuộc hàm số y=3x-7

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